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You will often be asked to find the $x$ -value when you are given a $y$ -value, or find the $y$ -value when you are given an $x$ -value. To find $x$ -values and $y$ -values, follow these two steps:

Rule

Finding $x$ -values and $y$ -values

To find the $y$ -value:: Insert the $x$ -value you have been given into the function, and calculate the value.
To find the $x$ -value:: Set $f (x)$ equal to the $y$ -value you’ve been given, and solve the equation for $x$ .

Example 1

You have the function $f (x) = 2 x - 3$ . Find the $x$ -value when $f (x) = 2$ , and the $y$ -value when $x = 2$ .

To find $x$ when $f (x) = 2$ :

\begin{array}{c} f (x) & = 2 \\ 2 x - 3 & = 2 \\ 2 x & = 2 + 3 \\ 2 x & = 5 & | \div 2 \\ x & = 2.5 . \end{array}

\begin{array}{c} f (x) = 2 x - 3 & = 2 \\ 2 x & = 2 + 3 \\ 2 x & = 5 & | : 2 \\ x & = 2.5 . \end{array}

Therefore, $x = 2.5$ when $f (x) = 2$ . To find $f (x)$ when $x = 2$ :

\begin{array}{c} f (2) & = 2 \cdot 2 - 3 \\ = 4 - 3 \\ = 1 . \end{array}

f (2) = 2 \cdot 2 - 3 = 4 - 3 = 1 .

Therefore,

f (x) = 1

when

x = 2

Note! To figure out where

f (x)

intersects the axes, this is the procedure:

1.: To find where $f (x)$ intersects the $y$ -axis, you insert 0 for $x$ . So you calculate $f (0)$ , and the answer is the point of intersection between the graph and the $y$ -axis, $(0, f (0))$ .
2.: To find where $f (x)$ intersects the $x$ -axis, you set $f (x) = 0$ . Solve the equation for $x$ , and the answer is the point of intersection between the graph and the $x$ -axis. These points are also called zeros of the function.

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A function works such that you end up with a relationship between variables. It works just like a machine. You put something in, and something new comes out.

Imagine a bread maker machine. If you put in finely ground, bleached flour, you end up with a smooth, white loaf. If you put in seeded, rough flour, you get multigrain wheat bread. The difference between a function and a bread maker is that you put numbers into a function—not flour!

Two examples of functions are $y = x + 2$ and $f (x) = x^{2} - 2$ . The function $y$ is the straight line in Figure (b), and $f (x)$ is the graph in Figure (d) below.

As you can see from the two functions in this section, one begins with $y$ and the other with $f (x)$ (read as “ $f$ of $x$ ”). Why? Functions can have different names, some of the most common ones are $y$ and $f (x)$ . Both tell you that they are values from the second axis.

The notation $f (x)$ tells you that you have a function that depends on the $x$ -value. That is, you put the numbers from the $x$ -axis into the function, and the value you get is a number on the $y$ -axis. Since $f (x)$ gives you $y$ -values, you can think of them as equals, so that $y = f (x)$ . A function thus receives a value, and returns a value.

You can insert many $x$ -values into a function, while the $y$ -values are directly dependent on what the $x$ -values are. We therefore call the $x$ -value the independent variable and the $y$ -value the dependent variable.

$\"A$

Theory

Definition of a Function

For each value of $x$ there is only one value of $y$ .

$\"Circle$

(a) Circle: Not a function

$\"Straight$

(b) Straight line = linear function

$\"Winding$

$\"Parabola$

(d) Parabola = quadratic function

Above you see four figures. Figures (b) and (d) show graphs that are functions. From the figures you can see that each

x

-value has only one corresponding

y

-value.

In Figures (a) and (c) you see that an $x$ -value can have several different $y$ -values. These are thus not functions.

In Figure (a), the $x$ -value we chose has two $y$ -values, which is true of almost any x value in a circle. In Figure (c) the $x$ -value we chose has four $y$ -values. We call these types of figures curves.

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\n \n \n \n \n $\"Situations$ \n

Functions are everywhere! It may sound a bit over the top, but functions are actually all over the place: all programmable things, all computer systems, mobile phones, apps, financial analyses, statistics, political analyses, price calculations, taxes, income calculations, earthquake warnings, landslide warnings, tsunami alerts, electronics, political forecasts, population growth forecasts, forecasts for sustainability—all forecasts in general! These are just a few examples where functions are used.

If you are going to study at a university, you are going to see functions. They are used in most subjects, including non-mathematical ones.

The reason is that functions say something about the connection or relationship between things. “What happens if $\dots$ ?” In this type of question, you can create mathematical models, and mathematical models are often functions. Given that you have a data set to be organized and analyzed, it is in this context you use functions. The function will then process the data set so that you can analyze the information.

Let’s consider Snapchat, Facebook or Instagram. All these apps are composed of functions. Some functions sort your friends list, other functions select filters. A third type of functions get used in the process of connecting your device to the internet. As you can see, functions make your life cooler, easier and more fun.

\n \n \n \n \n $\"Functions$ \n

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A function is like a machine. You put something in, and something else comes out. When we talk about functions in mathematics, we’re really talking about a connection between variables. This connection can be anything, but it follows a certain requirement. It’s this requirement that defines a function.

Before you move on, it is important that you know and understand the term argument.

An argument in the theory of functions is the number or the expression that you insert into the function. Up until now, you’ve probably thought of the argument as the $x$ in a function.

Theory

Definition of a Function

If at any given value of $x$ , there exists only one value of $y$ , you say that “ $y$ is a function $f$ of $x$ ”. You write $y = f (x)$ . In other words, the connection that meets the requirement of “only one value of $y$ to a given value of $x$ ” is a function.

Here is a figure of how you can imagine functions:

$\"x$

The graph of a function is the collection of points

(x, f (x))

in a coordinate system— in other words, the curve that passes through all the points.

You can use a function to:

Find $y$ -values when you have $x$ -values.

For example, to find the total price of the apples you are buying by weight, when you know the weight of the apples you have picked.

Find $x$ -values when you have $y$ -values.

For example, to find the weight of apples you are buying by weight, when you’re told the total price of the apples.

Find a connection between a set of $x$ -values and a set of $y$ -values.

Finding a function that shows a connection between two variables is called regression. For example, to find a function of the price of apples by their weight. You might have a list of the weights of different bags of apples, and a list of how much each bag costs.

Theory

Functional value and argument

$x$ is called the argument of the function. This is also called the independent variable. That is because you pick the values for $x$ . $y = f (x)$ is called the functional value of the function. This is also called the dependent variable. That is because it is the function that decides what the $y$ -value is when you have chosen the $x$ -values.

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A coordinate system is a grid that is formed by two number lines crossing each other at the point where the number zero appears on both lines. This intersection is called the origin. A coordinate system looks like this:

$\"The$

The $x$ -axis is also called the first axis, while the $y$ -axis is called the the second axis.

You will come across coordinate systems both in math at school and in daily life. Coordinate systems are a tool that help you know where things are located.

Think About This

Look at the coordinate system with the positive numbers on the figure. How do you think you read a point in the grid?

The answer is revealed further down on the page.

A point in the grid is called a coordinate. Every coordinate consists of one value from the first axis and one value from the second axis (see figure).

Below, you can see a coordinate system with points.These points are named after letters in the alphabet. This is how you find the coordinates of a point:

Rule

Read the Coordinates of a Point

1.: Put your finger on the point, then drag your finger straight down until you hit the first axis. Write down the number that is now under your finger.
2.: Put your finger back on the point, then drag your finger straight left until you hit the second axis. Write down the number that is now under your finger.
3.: The coordinates should look like this: $(x, y)$ . You have to replace the x with the number you found on the first axis. You have to replace the y with the number you found on the second axis.

$\"A$

Think About This

Can you read the coordinates of the different points above?

$A = (1, 1)$	$B = (3, 1)$	$C = (2, 3)$
$D = (4, 3)$	$E = (1, 5)$	$F = (3, 10)$
$G = (8, 7)$	$H = (5, 7)$	$I = (7, 4)$
$J = (7, 7)$	$K = (0, 4)$	$L = (6, 0)$

$A = (1, 1)$	$B = (3, 1)$	$C = (2, 3)$	$D = (4, 3)$
$E = (1, 5)$	$F = (3, 10)$	$G = (8, 7)$	$H = (5, 7)$
$I = (7, 4)$	$J = (7, 7)$	$K = (0, 4)$	$L = (6, 0)$

Math Vault

Want to solve exercises about the coordinate system? Try Math Vault!

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There are many different types of function tables. Here, you’ll look at two types: One that is nice in the beginning when you are learning, because it contains a lot of detail, and one that is a bit more concise and professional.

A function table is a table that shows which coordinates should be plotted in the coordinate system, so that you can draw the graph of the function.

A comprehensive function table has four columns: A column for the $x$ -values you need, a column for the calculation of each $y$ -value, a column where the $y$ -values that were calculated is written, and a column for the coordinates you get from the chosen $x$ -value and the corresponding $y$ -value so that you can plot the graph. It looks like this:

$\"Function$

In the first column, you place the $x$ -values that you want to examine. Why did I choose exactly these values? Well, which values you need follow from the problem you’re working on. Plus, you are often told which $x$ -values you should examine. If you can’t figure it out, you can start with values between $- 5$ and 5. This is a good place to begin.

In the second column, you see the function $y = 2 x + 3$ . This is a function of a straight line. In this column, you insert the chosen $x$ -values into the function and calculate the $y$ -values.

Then, you write the corresponding $y$ -values in the third column. If you calculate the contents in every cell in the second column, you see that the answers are exactly the values that are written in the third column.

Finally, in the fourth column, the $x$ -value and $y$ -value are put together into coordinates $(x, y)$ . These are the coordinates you have to insert in the coordinate system and draw lines through to make your graph.

Professional Function Table

The professional function table contains precisely what you need to find the coordinates, so that you can draw the graph. The difference from the previous function table is that Columns 2 and 4 are removed, and you write the table horizontally instead of vertically (changing columns into rows).

Even if you don’t include the coordinate column, you see that the $x$ -values and $y$ -values are included, and it is just these that make up the coordinates. In other words, $(x, y)$ is included, it just looks a little different.

$\"Professional$

Here’s the graph with the points marked:

$\"Graph$

Example 1

Lance is $30$ from home. He rides his bike at a constant speed of $12$ . Make a graphical illustration of the number of hours and the number of kilometers he is from home. How much time does it get before he gets home?

To solve this problem, you should first convert the information from the exercise into a function. The formula for a straight line is most suitable, since Lance is riding his bike at a constant speed.

Let’s examine how words and expressions are converted to mathematical symbols. When solving this problem you should think like this: Text informs us that Lance is $30$ km away from home and that he is riding his bike at $12$ km/h. So his journey begins at $30$ km distance, and the closer he gets to home, the shorter the distance. Thus, you convert the information like this:

$f (x)$ is the distance to home at any given time. This distance gets smaller and smaller, the longer Lance travels towards home.
$x$ is the number of hours Lance rides his bike.
$30$ km is the distance that is counted down to $0$ km, from where Lance starts, until he reaches home.
$12$ km/h is Lance’s speed.

Now you insert these items into the formula for a straight line, $f (x) = m x + b$ , and get:

f (x) = - 12 x + 30

30 is the constant term since the starting point for Lance is set— $30$ km from home. $- 12$ is the slope, since the distance to home is reduced by $12$ km for every hour Lance rides his bike towards home.

Note! The minus sign for Lance’s speed does not mean that he’s riding his bike in reverse all the way home! Instead, the minus sign is used to count down from $30$ km to $0$ km. The distance should decrease, so you need a minus sign to accomplish this.

Based on the expression above you get the following function table:


$x$	$y = f (x) = - 12 x + 30$	$(x, y)$

$0$	$f (0) = - 12 \cdot 0 + 30 = 30$	$(0, 30)$

$0.5$	$f (0.5) = - 12 \cdot 0.5 + 30 = 24$	$(0.5, 24)$

$1$	$f (1) = - 12 \cdot 1 + 30 = 18$	$(1, 18)$

$1.5$	$f (1.5) = - 12 \cdot 1.5 + 30 = 12$	$(1.5, 12)$

$2$	$f (2) = - 12 \cdot 2 + 30 = 6$	$(2, 6)$

$2.5$	$f (2.5) = - 12 \cdot 2.5 + 30 = 0$	$(2.5, 0)$

$3$	$f (3) = - 12 \cdot 3 + 30 = - 6$	$(3, - 6)$

From the function table, you see that Lance gets home after $x = 2.5$ hours. Since the problem asks you to solve this graphically, you now have to make a coordinate system in which you mark the points you have calculated and written in the column with $(x, y)$ , and then draw the line between them. The graph should look like this:

$\"Example$

Lance arrives at home when the $y$ -coordinate is 0. This is when the graph intersects the $x$ -axis. The circle marks this point and you can see that Lance arrives at home after $x = 2.5$ hours. Thus, the answer is:

It takes $2.5$ hours for Lance to arrive at home.

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You will often be asked to find the $x$ -value when you are given a $y$ -value, or find the $y$ -value when you are given an $x$ -value. To find $x$ -values and $y$ -values, follow these two steps:

Rule

Finding $x$ -values and $y$ -values

To find the $y$ -value:: Insert the $x$ -value you have been given into the function, and calculate the value.
To find the $x$ -value:: Set $f (x)$ equal to the $y$ -value you’ve been given, and solve the equation for $x$ .

Example 1

You have the function $f (x) = 2 x - 3$ . Find the $x$ -value when $f (x) = 2$ , and the $y$ -value when $x = 2$ .

To find $x$ when $f (x) = 2$ :

\begin{array}{c} f (x) & = 2 \\ 2 x - 3 & = 2 \\ 2 x & = 2 + 3 \\ 2 x & = 5 & | \div 2 \\ x & = 2.5 . \end{array}

\begin{array}{c} f (x) = 2 x - 3 & = 2 \\ 2 x & = 2 + 3 \\ 2 x & = 5 & | : 2 \\ x & = 2.5 . \end{array}

Therefore, $x = 2.5$ when $f (x) = 2$ . To find $f (x)$ when $x = 2$ :

\begin{array}{c} f (2) & = 2 \cdot 2 - 3 \\ = 4 - 3 \\ = 1 . \end{array}

f (2) = 2 \cdot 2 - 3 = 4 - 3 = 1 .

Therefore,

f (x) = 1

when

x = 2

Note! To figure out where

f (x)

intersects the axes, this is the procedure:

1.: To find where $f (x)$ intersects the $y$ -axis, you insert 0 for $x$ . So you calculate $f (0)$ , and the answer is the point of intersection between the graph and the $y$ -axis, $(0, f (0))$ .
2.: To find where $f (x)$ intersects the $x$ -axis, you set $f (x) = 0$ . Solve the equation for $x$ , and the answer is the point of intersection between the graph and the $x$ -axis. These points are also called zeros of the function.

When working with functions, you often need to provide $x$ -values and $y$ -values in your answer. In mathematics, the term domain is used for the set of $x$ -values that fit in the problem, and range is used for the corresponding set of $y$ -values. Formally, they are referred to as this:

Theory

Domain and Range

The set of all $x$ -values that fit in a function $f$ is called the domain of $f$ and is written $D_{f}$ . The set of all $y$ -values determined by $D_{f}$ is called the range of $f$ and is written $R_{f}$ .

These are the most common examples:

For linear functions, the range is the set of all real numbers. You write $R_{f} = ℝ$ .
For quadratic functions, the range is either all numbers from the minimum and upwards, or all numbers from the maximum and downwards. You then write either $R_{f} =$ or $R_{f} =$ , where $a$ is the $y$ -value of the extremum.

For rational functions expressed as

f (x) = \frac{a x + b}{c x + d},

the range is all numbers except the $y$ -value where the denominator is zero, that is, the horizontal asymptote. You can write it as

R_{f} = ℝ ∖

since $\frac{a}{c}$ is the horizontal asymptote.

Example 1

Robert is on land and throws a rock into the water. The stone follows a path given by the function $f (x)$ , where $x$ is the distance from Robert to the stone in the horizontal direction, and $y$ is the height the stone has above the water at all times. You will learn that the height of the stone follows the function

f (x) = - 0.1 x^{2} + x + 2

Find the domain $D_{f}$ and the range $R_{f}$ for the function so that the throw makes sense.

To find the domain $D_{f}$ you must find the smallest and largest value that $x$ can have. You see that $x$ must be greater than 0. If you allow $x$ to be less than zero, Robert will end up throwing the stone further ashore and that does not match the exercise.

You must now find the largest value of $x$ . The throw ends when the stone hits the water surface and that happens when $y = 0$ . You must therefore solve the equation

f (x) = 0 .

Since $f (x)$ is a quadratic function you can use the quadratic formula to solve this equation. The solutions are therefore $x_{1} = - 1.71$ and $x_{2} = 11.71$ . Since the domain starts at 0, you can ignore $x_{1}$ since $x_{1}$ is negative. The stone will hit the water after $11.71$ meters. The domain is therefore

D_{f} =

To find the range $R_{f}$ , you need to find the smallest and largest value $y$ can be from the $x$ -values you found for $D_{f}$ . The smallest $y$ -value is when the rock hits the water surface. This happens at $y = 0$ — that is, at 0 meters above the water.

To find the largest $y$ -value, you must find the $y$ -value when the rock is at its greatest height above the water surface. This can be found by reading the $y$ -value at the maximum of the function.

You can find the maximum by drawing the function or by using derivation. Regardless of the method you choose, the greatest height the stone will be is $4.5$ m. The range is then

R_{f} =

Theory

Asymptotes, Domain and Range

A vertical asymptote intersects the $x$ -axis. The mathematical way of writing a vertical asymptote is $x = a$ , where $a$ is the number where the vertical asymptote intersects the $x$ -axis. The domain of the function with a vertical asymptote does not contain $a$ .
A horizontal asymptote intersects the $y$ -axis. The mathematical way of writing a horizontal asymptote is $y = a$ , where $a$ is the number where the horizontal asymptote intersects the $y$ -axis. The range of the function with a horizontal asymptote does not contain $a$ .

Example 2

Find the domain and the range of $f (x) = \frac{2 x + 3}{x - 1}$

You can find the domain $D_{f}$ by looking at the $x$ -values where the function makes sense, that is, where you can insert an $x$ -value and get an answer. You know that you can’t have zero in the denominator of a fraction. For this reason the function is not defined for the $x$ -value that makes the denominator 0. Therefore you get

x - 1 = 0 \Leftrightarrow x = 1 .

The domain is then all values of $x$ except the $x$ -value of the vertical asymptote, in this case $x = 1$ . You write this mathematically

D_{f} = ℝ ∖

You find the range $R_{f}$ by finding the horizontal asymptote. That is to say, the range is all real numbers except the horizontal asymptote. Since $f$ describes a hyperbola (as you can see from the function), you find the horizontal asymptote using the formula

y = \frac{a}{c} .

Then you get

y = \frac{2}{1} = 2 .

The range is then: All real numbers except the $y$ -value of the horizontal asymptote, in this case $y = 2$ . You write this mathematically as

R_{f} = ℝ ∖

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An intersection point between two functions is a point where the graphs of the functions meet each other. You find intersections by solving the system of equations consisting of the two functions.

Rule

The Intersection Between Two Graphs

You find the point where the graph of $f$ and the graph of $g$ intersects by solving the equation $f (x) = g (x)$ .

Example 1

Find the intersection between $f (x) = - x - 1$ and $g (x) = 2 x + 8$

Since $f (x) = y$ and $g (x) = y$ , you can set them equal to each other:

\begin{array}{c} - x - 1 & = 2 x + 8 \\ - 3 x & = 9 \\ x & = - 3 \end{array}

Insert $x$ into $f (x) = - x - 1$ because it’s the simpler expression of the two. You could also insert $x$ into $g (x)$ , but it would be more work.

f (- 3) = - (- 3) - 1 = 2 .

The intersection between $f (x)$ and $g (x)$ is

(x, y) = (- 3, 2) .

$\"Intersection$

Example 2

Find the intersection between $f (x) = x^{2} + 3 x - 2$ and $g (x) = 2 x + 3$

Since $f (x) = y$ and $g (x) = y$ , you can set them equal to each other:

\begin{array}{c} x^{2} + 3 x - 2 & = 2 x + 3 \\ x^{2} + x - 5 & = 0 \end{array}

Solving the quadratic equation with the quadratic formula:

\begin{array}{c} x & = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 5)}}{2 \cdot 1} \\ = \frac{- 1 \pm \sqrt{1 + 20}}{2} \\ = \frac{- 1 \pm \sqrt{21}}{2}, \end{array}

\begin{array}{c} x_{1} & = \frac{- 1 - \sqrt{21}}{2} \approx - 2.79, \\ x_{2} & = \frac{- 1 + \sqrt{21}}{2} \approx 1.79 . \end{array}

Insert $x_{1}$ and $x_{2}$ into $g (x) = 2 x + 3$ because it’s the simpler expression. You could also insert $x_{1}$ and $x_{2}$ into $f (x)$ , but it would be more work.

\begin{array}{c} y_{1} & = g (- 2.79) = 2 \cdot (- 2.79) + 3 = - 2.58 \\ y_{2} & = g (1.79) = 2 \cdot 1.79 + 3 = 6.58 \end{array}

\begin{array}{c} y_{1} & = g (- 2.79) = 2 \cdot (- 2.79) + 3 = - 2.58 \\ y_{2} & = g (1.79) = 2 \cdot 1.79 + 3 = 6.58 \end{array}

The intersection between

f (x)

and

g (x)

\begin{array}{c} (x_{1}, y_{1}) & = (- 2.79, - 2.58), \\ (x_{2}, y_{2}) & = (1.79, 6.58) . \end{array}

$\"f(x)$

Example 3

For which values of $x$ are $f (x) = \sin x$ and $g (x) = \cos x$ equal?

Find the point of intersection between the graphs by setting them equal to each other:

\begin{array}{c} \sin x & = \cos x \\ \tan x & = 1 \\ x & = \tan^{- 1} (1) = \frac{π}{4} + n π \end{array}

You need to check that you didn’t miss any solutions since you divided by $\cos x$ , which can be 0. You check this by looking at what happens if $\cos x = 0$ . Then $x = \frac{π}{2}$ , which means $\sin x = 1$ , or $x = \frac{3 π}{2}$ , which means $\sin x = - 1$ . In both cases, $\sin x$ is different from $\cos x$ , so they are not solutions.

There are infinitely many points of intersection, and the points have $x$ -values equal to $x = \frac{π}{4} + n π$ for $n \in ℤ$ . You find the $y$ -values by inserting the $x$ -values into one of the functions, for example to $f (x) = \sin x$ . Here, you have to be aware that even if $\tan x$ has a period of $π$ , you have found two different angles on the unit circle, with $π$ radians between. This is why you get two different values when you put the result back into $f (x) = \sin x$ : $\begin{array}{c} f & = \sin \frac{π}{4} = \frac{\sqrt{2}}{2} \\ f & = \sin \frac{5 π}{4} = - \frac{\sqrt{2}}{2} \end{array}$

These are two different $y$ -values, $y = \frac{\sqrt{2}}{2}$ and $y = - \frac{\sqrt{2}}{2}$ , each belonging to their respective angle on the unit circle. These angles repeat themselves with a period of $2 π$ . Together, you get two different points of intersection:

\begin{array}{c} (x_{1}, y_{1}) & = \\ (x_{2}, y_{2}) & = \end{array}

\begin{array}{c} (x_{1}, y_{1}) & = \\ (x_{2}, y_{2}) & = \end{array}

Note! In this type of exercise, you are pleased with an answer containing

n

. This is because your were asked to find all the points, meaning the general solution, not the points within a certain interval.

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The zeros or roots of a function tell us where a graph intersects the $x$ -axis. Since we’re talking about intersection with the $x$ -axis, you know that $y = 0$ . That means that you can find the zeros by solving the equation $f (x) = 0$ .

Rule

Zeros

You find the zeros of a function by solving the equation

f (x) = 0 .

The sign chart of $f (x)$ tells you when the graph of $f$ is above or below the $x$ -axis, and where $f (x)$ intersects the $x$ -axis.

$\"Graph$

Example 1

Find the zeros of $f (x) = x^{2} + 5 x + 6$

You find the zeros by solving the equation $f (x) = x^{2} + 5 x + 6 = 0$ : $\begin{array}{c} x & = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 6}}{2} \\ = \frac{- 5 \pm \sqrt{25 - 24}}{2} \\ = \frac{- 5 \pm 1}{2} \end{array}$

This gives

\begin{array}{c} x_{1} & = \frac{- 5 - 1}{2} = - 3, \\ x_{2} & = \frac{- 5 + 1}{2} = - 2 . \end{array}

So the graph meets the $x$ -axis at the points i $(- 3, 0)$ and $(- 2, 0)$ .

Example 2

Find the zeros of $g (x) = 2 \sin (2 x) + 1$ in the interval $x \in$

Again, you can find the zeros by solving the equation $g (x) = 0$ : $\begin{array}{c} 2 \sin (2 x) + 1 & = 0 \\ \sin (2 x) & = - \frac{1}{2} \end{array}$

The basic equation has the solutions

\begin{array}{c} 2 x & = - \frac{π}{6} + 2 π n, \\ 2 x & = π - \end{array}

Solving these for $x$ , you get

\begin{array}{c} x & = - \frac{π}{12} + π n, \\ x & = \frac{7 π}{12} + π n . \end{array}

Since the interval is $x \in$ , you get that $x \in$ . The graph meets the $x$ -axis at the points and .

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a = 0

Formula

The Formula for Linear Functions

f (x) = a x + b,

where $a$ is the slope of the line and $b$ is called the constant term, and is the $y$ -coordinate where the graph intersects the $y$ -axis. The slope $a$ tells you how much the graph increases or decreases when the $x$ -value increases by 1.

$\"The$

Example 1

The function

f (x) = - 0.5 x + 2

tells you that the graph intersects the $y$ -axis in the point $(0, 2)$ and that the slope $a = - 0.5$ . That means that when $x$ increases by 1, $y$ decreases by $0.5$ . The graph is therefore sloping downwards from left to right. It looks like this:

$\"Example$

Rule

Important Properties of the Linear Function

The slope $a$ tells you how much the graph increases/decreases when $x$ increases by 1.
If $a > 0$ , then the graph slopes upwards from left to right, and if $a < 0$ , it slopes downwards.
The graph intersects the $y$ -axis at the point $b$ .
The graph is a straight line with coordinates $(x, y) = (x, f (x))$ .

Example 2

Chelsea Clinton had a babysitter when she was a kid. Her father, Bill, was a bit stingy and decided to pay the babysitter $$ 7.50$ per hour plus $$ 5$ for travel to their home. Find a function that shows how much Bill had to pay the babysitter for $x$ hours.

You know that Bill has to pay $ $5$ every time the babysitter comes to take care of Chelsea. You also know that he pays $ $7.50$ per worked hour. The babysitter thus gets $ $7.50$ for one hour, $7.50 \cdot $ 2 = $ 15$ for two hours, and so on. Bill pays thus $7.50 x$ dollars when the babysitter works $x$ hours. The expression for how much Bill pays the babysitter looks like this:

y = 7.5 x + 5

How much does Bill have to pay for the afternoon when he and Hillary have to go to a meeting at the White House and need a babysitter for 3 hours?

Here $x = 3$ , so you get the following calculation:

y = 7.5 \cdot (3) + 5 = $ 27.50

The babysitter gets $ $27.50$ for 3 hours of work.

The babysitter planned to go to a U2 concert a few hours after she had to babysit. Tickets for the concert cost $$ 35$ . She forgot her wallet at home, and hoped that she would earn enough money during the afternoon to cover the ticket. How late do Bill and Hillary have to be in order for the babysitter to have enough money for the concert?

Here you know that the babysitter must earn at least $ $35$ to be able to afford the ticket, meaning $y = 35$ . What you need to find is how long she must babysit— that is, a value for $x$ . You put in the value for $y$ and solve the equation for $x$ : $\begin{array}{c} 35 & = 7.5 x + 5 \\ 35 - 5 & = 7.5 x \\ 30 & = 7.5 x & | \div 7.5 \\ x & = 4 \end{array}$

To be able to afford the concert, she had to babysit for four hours. So she hoped that Bill and Hillary were $4 - 3 = 1$ hour late. If they were not, she would have to borrow money from a friend, or ask Bill for a pay rise.

You need two points to find the function of a line. If you know two points that are on the graph, you can use the two formulas below to find the slope $a$ and the constant term $b$ . The symbol $Δ$ (delta) is a Greek letter. In mathematics you use it to describe change or difference. This means that you read the formula below as “ $a$ is equal to the change in $y$ divided by the change in $x$ ”.

Formula

How to Find a Function From Two Points

The line through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ has the slope

a = \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}},

where $y_{1} = f (x_{1})$ and $y_{2} = f (x_{2})$ . Note! Study the figure closely!

$\"Linear$

To find the constant term

b

, you need to use the following formula:

b = y_{1} - a x_{1} .

Example 3

Find the slope of the line that goes through the points $(5, 2)$ and $(3, 6)$ , and its intersection with the $y$ -axis

You let $(x_{1}, y_{1}) = (3, 6)$ and $(x_{2}, y_{2}) = (5, 2)$ . You would get the same result if you swap $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . By inserting those values into the formula, you get

a = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 6}{5 - 3} = \frac{- 4}{2} = - 2 .

You now know that your line decreases by 2 when you move one place to the right. Let’s see what the intersection with the $y$ -axis is:

\begin{array}{c} b & = y_{1} - a x_{1} \\ = 6 - (- 2) \cdot 3 \\ = 6 + 6 \\ = 12 \end{array}

\begin{array}{c} b = y_{1} - a x_{1} = 6 - (- 2) \cdot 3 = 6 + 6 = 12 \end{array}

The intersection with the

y

-axis is then

(0, 12)

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f (x) = a x + b

. Often,

a

represents a unit price, while

b

represents a fixed price.

Example 1

Find an expression for the price of a BASE jump insurance policy with a fixed price of $$ 1000$ and a price of $$ 230$ per jump

From the text you can see that there is an amount that is fixed regardless of how many jumps you do, and an amount that varies with the number of jumps. If you set $x$ equal to the number of jumps, the function looks like this:

f (x) = 230 x + 1000 .

The price is 230 times the number of jumps plus the fixed price.

Rule

How to Find a Linear Function

$a$ is the amount that varies,

$b$ is the amount that occurs one time.

Example 2

Let’s say you are old enough to have your driver’s license. You are going on a trip and have to rent a car. You are considering two types of leases.

Lease agreement “Smart Young 1” is such that you pay a fixed amount of $$ 24.90$ , and then $$ 7$ per 10 miles you drive the car.

Lease agreement “Smart Young 2” is such that you pay a fixed amount of $$ 29.90$ , and then $$ 5$ per 10 miles you drive. Which lease should you choose if you plan to drive a trip of 60 miles?

In this case, you need to find a function for each lease, then plot them in the same coordinate system. Then you need to make an assessment. In this case, let $x$ represent 10 miles, so that $x = 1$ is 10 miles, $x = 2$ is 20 miles, and so on.

Lease “Smart Young 1”

The fixed price is $ $25.00$ , and the cost per mile is $ $7$ . This gives you the function

f (x) = 7 x + 25

Lease “Smart Young 2”

The fixed price is $ $30.00$ , and the cost per mile is $ $5$ . This gives you the function

g (x) = 5 x + 30

You plot $f$ and $g$ and get these graphs:

$\"Example$

You can see that the two lines intersect at

x

2.5

. Remember that

x

represents 10 miles.

That means that lease “Smart Young 2” is cheaper if you drive over 25 miles, and “Smart Young 1” is cheaper if you drive less than 25 miles. They cost the same if you drive exactly 25 miles.

"},"siblings":[{"h1":"What Are Linear Functions?","path":"/encyclopedia/functions/theory-of-functions/linear-functions/what-are-linear-functions"},{"h1":"How Are Linear Functions Used in Real Life?","path":"/encyclopedia/functions/theory-of-functions/linear-functions/how-are-linear-functions-used-in-real-life"},{"h1":"What Makes a Function Proportional?","path":"/encyclopedia/functions/theory-of-functions/linear-functions/what-makes-a-function-proportional"}],"key":"content/en/encyclopedia/05 Functions/01 Theory of Functions/02 Linear Functions/202FunctionsInRealLife.tex"},"level":"11","metaDescription":"Applying math to create functions from a practical problem is one of the most useful skill you can learn. Learn this skill by studying this entry.","metaDescriptionActivity":null,"activities":null},{"assets":null,"name":"What Makes a Function Proportional?","howContent":null,"encyclopediaArticle":{"htmlContent":{"html":"

You often come across two quantities whose relationship is such that if you double one of them, the other one doubles too. For example, if you are baking, and you need $1$ kg flour to make 24 buns, it is easy to figure out how much flour you need to make twice as many buns. You need $2$ kg flour to make 48 buns. Doubling one quantity leads to the other one doubling!

The mathematical word for this concept is proportionality. If all values of $y$ divided by all corresponding values of $x$ give you a certain constant $k$ , that means $x$ and $y$ are proportional quantities. We call this constant $k$ the proportionality constant. We usually write this as shown in the rule box below— $y$ is $k$ multiplied by $x$ . The two explanations below for why are exactly the same:

Rule

A calculation shows that the expressions are equal:

\begin{array}{c} \frac{y}{x} & = k \\ x \cdot \frac{y}{x} & = k \cdot x \\ y & = k x \end{array}

Rule

Two entities $x$ and $y$ are proportional if

y = k x,

where $k$ is a constant.

You can see in Example 2 below what happens when $k = 1$ . Here are some rules for remembering what happens to different values of $k$ :

If the value of $k$ is higher than $1$ , then the line will be steeper.
If the value of $k$ is lower than $1$ , then the line will be more gradual.
If $k$ is a negative number, then the graph will slope downwards from left to right and eventually cross under the $x$ -axis.

A proportional function is a special case of the linear function $y = a x + b$ . If you set $a = k$ and $b = 0$ , then you get the formula of proportionality. The graph is always a straight line through the origin at $(0, 0)$ . The ratio between $y$ and $x$ is always equal to the slope $k$ .

Example 1

This graph shows

y = x

, which means that

k = 1

. The function is proportional, so if you take all the

y

-coordinates and divide it by the

x

-coordinates, the answer will always be

k = 1

Example 2

Is the function $y = \frac{2 x}{3}$ proportional?

You can find out by doing a few modifications:

\begin{array}{c} y & = \frac{2 x}{3} \\ = \frac{2 \cdot x}{3 \cdot 1} \\ = \frac{2}{3} \cdot \frac{x}{1} \\ \approx 0.67 \cdot x \approx 0.67 x \end{array}

y = \frac{2 x}{3} = \frac{2 \cdot x}{3 \cdot 1} = \frac{2}{3} \cdot \frac{x}{1} \approx 0.67 \cdot x \approx 0.67 x

You determined that

k \approx 0.67

k = \frac{2}{3}

, so the function is proportional.

Example 3

You are given the following coordinates:

$\"Function$

$\"Function$

Do the points follow a proportional function?

You know that if the points follow a proportional function, you’ll get the same answer when you divide the value of $y$ by the value of $x$ for all the coordinates:

\begin{array}{c} \frac{7}{1} & = 7 & \frac{14}{2} & = 7 \\ \frac{21}{3} & = 7 & \frac{28}{4} & = 7 \\ \frac{35}{5} & = 7 \end{array}

Since all the answers are equal, you have proportionality.

Note! Functions in the form $y = \frac{k}{x}$ are called inversely proportional.

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A polynomial function where the highest exponent is 2 is called a quadratic function. The graph of a quadratic function is called a parabola. It looks either like a happy smile, or a sad frown. You can see different parabolas below.

Theory

Quadratic Function

The quadratic function looks like this:

f (x) = a x^{2} + b x + c,

where $a$ , $b$ and $c$ are constants. The constants $a$ and $b$ are called coefficients. The term $a x^{2}$ is called a second-degree term or quadratic term, the term $b x$ is called a first-degree term or linear term, and $c$ is called the constant term.

Below you see a picture of the graphs of four different quadratic functions. Note that all have either a maximum or minimum.

$\"Graph$

Rule

Parabola (Maximum and Minimum)

When $a > 0 (positive) \Rightarrow ⋃$ : The graph is smiling, and the function has a minimum (vertex).
When $a < 0 (negative) \Rightarrow ⋂$ : The graph is sad, and the function has a maximum (vertex).

Example 1

You have the quadratic function

f (x) = 0.5 x^{2} - x - 3

Determine if the parabola has a maximum or a minimum.

You see that the coefficient in front of $x^{2}$ is a positive number ( $a > 0$ ). Therefore, the graph smiles and have a minimum.

$\"Example$

Example 2

You have the quadratic function

f (x) = - x^{2} - x + 12

Determine if the parabola has a maximum or a minimum.

In this case, the coefficient in front of $x^{2}$ is a negative number ( $a < 0$ ). Therefore, the graph will face downwards and have a maximum, as in the figure below:

$\"Example$

There are several methods to use to find the maximum or minimum. You can use the derivative, a sign chart, or the method that follows here:

Rule

Finding the Maximum or Minimum of a Parabola

When you have the quadratic function

f (x) = a x^{2} + b x + c,

you can find the $x$ -values and $y$ -values of the vertex like this:

The $x$ -value of the maximum or minimum point:

x = - \frac{b}{2 a}

$y$ -value of the maximum or minimum point:

y = f

Example 3

Determine if the graph of $f (x) = x^{2} - 4 x + 4$ has a maximum or a minimum, and find this point

The function $f (x)$ has $a = 1$ , $b = - 4$ and $c = 4$ . Since $a = 1 > 0$ in the expression, you know that the graph is smiling and you have a minimum.

First, find the $x$ -value, and then the $y$ -value, of this minimum:

\begin{array}{c} x & = \frac{- (- 4)}{2 \cdot 1} = \frac{4}{2} = 2 \\ y & = f (2) \\ = 2^{2} - (4 \cdot 2) + 4 \\ = 4 - 8 + 4 \\ = 0 \end{array}

\begin{array}{c} x & = \frac{- (- 4)}{2 \cdot 1} = \frac{4}{2} = 2 \\ y & = f (2) = 2^{2} - 4 \cdot 2 + 4 = 4 - 8 + 4 = 0 \end{array}

The minimum is then

(2, 0)

. Upon inspection of

f (x) = x^{2} - 4 x + 4

, you can see that this is a square and therefore has only one zero. In this case, the zero and the minimum are the same point.

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A polynomial is an expression where all the terms have $x$ as the base, and where the exponents are non-negative integers (the exponents can be 0, 1, 2, $\dots$ ). The largest exponent of the expression determines the degree of the polynomial.

Theory

Polynomials

A polynomial of degree $n$ looks like this:

\begin{array}{c} a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + \dots \end{array}

\begin{array}{c} a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + \dots + a_{2} x^{2} + a_{1} x + a_{0} \end{array}

Note! Remember that constants, which don’t appear at first glance to have an $x$ in them, actually have an $x$ raised to the power zero – $x^{0}$ – which equals $1$ . A polynomial function is a function where $f (x)$ is set equal to a polynomial. Linear functions and quadratic functions are the most common polynomial functions. Linear functions are degree 1 polynomials and quadratic functions are degree 2 polynomials. You may also encounter polynomial functions of a higher degree than this. Below, you can see a picture of five different polynomial functions with degrees from 2 to 7. For example, the pink graph is a quadratic function and the gray is a function of third degree.

$\"Graph$

One can usually see what degree a polynomial has by looking at its graph. A quadratic polynomial has one “bend”, a cubic (third degree) polynomial has two “bends”, a quartic (fourth degree) polynomial has three “bends”, etc.

But note that in some special cases, these “bends” are at the same point—and then you won’t see all of them.

Example 1

$f (x) = 2 x^{3} - 4 x + 3$ is a polynomial of degree 3 (a cubic polynomial).

$f (x) = 4 x^{2} - 3 x^{3} + x^{5}$ is a polynomial of degree 5 (a quintic polynomial).

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When something increases or decreases by the same percentage over a fixed period, you have an exponential (percentage) growth. The exponential growth may be negative, meaning the graph will decrease to the right rather than ascending upwards as it otherwise would.

Theory

Exponential Functions

An exponential function is expressed in this way:

f (x) = a \cdot b^{x}

Notice that the variable is now the exponent. Both $a$ and $b$ are numbers. We call $a$ the start value and $b$ the growth factor.

When the value of $a$ in the function is positive, the graph looks like one of the two graphs below.

Rule

The Graph of an Exponential Function

$\"The$

$a$ is the $y$ -value when $x = 0$ , $b$ is the growth factor,

$0 < b < 1$ blue graph, $b > 1$ red graph.

In general, $b > 1$ gives you a fixed percentage increase, $0 < b < 1$ gives you a fixed percentage reduction and $b = 1$ gives no change. The number $b$ acts as a growth factor. The value of $a$ affects the sign of the function values.

Example 1

Suppose you deposit $$ 5000$ into a savings account. You get $3$ interest per year on this deposit. How much will you have in your account after 7 years?

This is an example of exponential growth. First you need to find the growth factor associated with a $3$ % increase:

\begin{array}{c} Growth Factor & = 1 + \frac{3}{100} \\ = 1 + 0.03 \\ = 1.03 \end{array}

Growth Factor = 1 + \frac{3}{100} = 1 + 0.03 = 1.03 .

The initial value is

$ 5000

and the growth factor is

1.03

. Since the money is in the account for 7 years, you must therefore multiply the growth factor by itself seven times. In mathematics, this is written as a power with 7 as the exponent.

You must calculate the following terms to find the amount of money you have after seven years:

5000 \cdot 1.0 3^{7} = 6149.37 .

Thus, you have $$ 6149.37$ in your account after seven years.

Example 2

You have the function $f (x) = 3 \cdot 2^{x}$ . This function intersects the $y$ -axis at $y = 3$ and grows exponentially. This form of growth is very powerful. References to this type of graph are also often used in everyday speech, when someone says something is experiencing “exponential” growth.

$\"Example$

Example 3

You have the function $f (x) = 3 \cdot 0 . 5^{x}$ . This intersects the $y$ -axis at $y = 3$ and decreases exponentially. This form of reduction is very powerful. It’s like the downward version of the previous example.

$\"Example$

"},"image":{"_key":"ba490f04-4774-5167-9e7f-d82743dee94a","type":"px","dimension":"width","value":300,"url":"https://images.houseofmath.com/overleaf-assets/images/exponentialfunctionsgrowthfactor.svg","alt":"The graphs of an increasing and a decreasing exponential function"},"siblings":[{"h1":"What Are Quadratic Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-quadratic-functions"},{"h1":"What Are Polynomial Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-polynomial-functions"},{"h1":"What Are Exponential Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-exponential-functions"},{"h1":"Exponential Functions with Euler's Number","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/exponential-functions-with-eulers-number"},{"h1":"What Does Inversely Proportional Mean in Math?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-does-inversely-proportional-mean-in-math"},{"h1":"What Are Power and Root Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-power-and-root-functions"},{"h1":"What Are Logarithmic Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-logarithmic-functions"},{"h1":"What Are Rational Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-rational-functions"}],"title":"What Are Exponential Functions?","path":"/encyclopedia/functions/theory-of-functions/types-of-functions/what-are-exponential-functions","taskReference":null,"isRootPath":false,"sharedLanguageKey":null},"level":"15","metaDescription":"A constant growth factor leads to exponential growth. This growth is described by an exponential function. Learn about exponential functions here.","metaDescriptionActivity":null,"activities":null,"assets":null,"name":"What Are Exponential Functions?","howContent":null},{"level":"16","metaDescription":"Exponential function can have an arbitrary number as base, often the Euler's number e is used. Learn how to use exponential function and the base e.","metaDescriptionActivity":null,"activities":null,"assets":null,"name":"Exponential Functions with Euler's Number","howContent":null,"encyclopediaArticle":{"path":"/encyclopedia/functions/theory-of-functions/types-of-functions/exponential-functions-with-eulers-number","breadcrumbs":[{"path":"/encyclopedia","h1":"Encyclopedia","_key":"c0cf1a4b-5f7d-5390-9a52-e6ae2428b50c"},{"_key":"6c073836-cdf3-5d60-96fc-ac4946f467f6","path":"/encyclopedia/functions","h1":"Functions"},{"_key":"a591aa17-e072-5085-be73-37f2d8459442","path":"/encyclopedia/functions/theory-of-functions","h1":"Theory of Functions"},{"path":"/encyclopedia/functions/theory-of-functions/types-of-functions","h1":"Types of Functions","_key":"54ec2c96-57ab-5746-b8fe-fbb057ed55e3"},{"path":"/encyclopedia/functions/theory-of-functions/types-of-functions/exponential-functions-with-eulers-number","h1":"Exponential Functions with Euler's Number","_key":"229ff283-efbf-5249-b5da-45c8dad333f1"}],"isRootPath":false,"title":"Exponential Functions with Euler's Number","type":"encyclopedia","language":"en","htmlContent":{"html":"

When something increases or decreases by the same percentage in each period, you have exponential (percentage) growth. The exponential growth may be negative, meaning the graph decreases to the right rather than ascending upwards as it otherwise would.

In exponential functions, you can either use an arbitrary number $b$ as the base or $e$ as the base.

Theory

Exponential Functions

Exponential functions can have $e$ as the base or an arbitrary number $b$ as the base. In both cases, $a$ is a constant. They look like this:

f (x) = a \cdot e^{k x}

Note! These functions are reformulations of each other, so they have identical graphs . Note that the variable $x$ is now in the exponent! The symbols $a$ , $b$ and $k$ are numbers.

When the value of $a$ in the function is positive, the graph looks like one of the two graphs below.

Rule

The Graph of an Exponential Function

$\"The$

$a$ is the $y$ value when $x = 0$ , $b = e^{k}$ is the growth factor.

$0 < b < 1$ blue graph, $b > 1$ red graph.

Note! You are expected to be able to convert from $b = e^{k}$ and from $e^{k} = b$ .

Rule

Conversion from $b = e^{k}$ and from $e^{k} = b$

\begin{array}{c} b^{x} & = \\ e^{k x} & = \end{array}

The following is an overview of how the function behaves for different values of $a$ and $b > 0$ .

$a > 0$ and $b > 1$ :: The graph runs along the $x$ -axis and rises sharply to the right.

$\"Exponential$
$a > 0$ and $0 < b < 1$ :: The graph decreases sharply to the right and flattens along the $x$ -axis.

$\"Exponential$
$a \in ℝ$ and $b = 1$ :: The graph is a horizontal line through $a$ .
$\"Exponential$
$a < 0$ and $b > 1$ :: The graph runs along the $x$ -axis and decreases sharply below the $x$ -axis.

$\"Exponential$
$a < 0$ and $0 < b < 1$ :: The graph rises sharply and flattens along the $x$ -axis.

$\"Exponential$

In general, $b > 1$ gives you a fixed percentage increase, $0 < b < 1$ gives you a fixed percentage reduction, and $b = 1$ gives you no change.

The number $b = e^{k}$ acts as a growth factor. The value of $a$ affects the sign of the functional values.

Example 1

You have the function $f (x) = 3 \cdot 2^{x}$ . This intersects the $y$ -axis at $y = 3$ and grows exponentially. This form of growth is very powerful. References to this type of graph are also often used in everyday speech, when someone says something is experiencing “exponential” growth.

$\"Example$

Example 2

You have the function $f (x) = 3 \cdot 0 . 5^{x}$ . This intersects the $y$ -axis at $y = 3$ and decreases exponentially. This form of reduction is also very powerful. It’s like the downward version of the previous example.

$\"Example$

Example 3

In a particular chemical reaction, the concentration of a substance is given by

f (t) = 2.50 - 2.50 \cdot e^{- 0.012 t},

where $t$ is the time measured in seconds, and $f (t)$ is measured in mmol/L.
You are given the following tasks:

1.
What is the concentration after 15 seconds? How long does it take before the concentration is $2.00$ ?
2.
Draw the graph of $f$ . What value will the concentration approach if the reaction lasts for a very long time?
3.
What is the reaction rate when the concentration is $2.00$ ?

1.

To find the concentration after 15 seconds, you have to insert

t = 15

f (t)

. You then get

\begin{array}{c} f (15) & = 2.50 - 2.50 \cdot e^{- 0.012 \cdot 15} \\ \approx 0.412 \end{array}

\begin{array}{c} f (15) = 2.50 - 2.50 \cdot e^{- 0.012 \cdot 15} \approx 0.412 \end{array}

The concentration after 15 seconds is

0.412

mmol/L.

Now you need to find out how long it takes before the concentration is $2.00$ mmol/L. Here you have to solve the equation $f (t) = 2.00$ . You set it up and get

2.50 - 2.50 \cdot e^{- 0.012 t} = 2.00

You subtract $2.50$ on both sides, and get

- 2.50 \cdot e^{- 0.012 t} = - 0.50

Then you divide by $- 2.50$ , and the equation becomes

e^{- 0.012 t} = 0.2

You now have an equation in the form $e^{a} = b$ . You apply $\ln$ on both sides, and you get

- 0.012 t = \ln 0.2 = - 1.61

Finally, you divide by $- 0.012$ and find the solution

t = \frac{- 1.61}{- 0.012} \approx 134

This means that it takes 134 seconds before the concentration is $2.00$ mmol/L

2.

First you draw the graph of

f

. It looks like this:

$\"Example$

To find out which value the concentration is approaching if the reaction lasts for a very long time, you can either look at the graph and see that it goes towards $2.50$ , or insert a properly large $t$ and get $f (100$ . This means that if the reaction takes a very long time, the concentration approaches $2.50$ mmol/L Another way you can say this is to say that $y = 2.5$ is the horizontal asymptote.

3.

To find the reaction rate, you must differentiate the reaction function. You must compute

f^{'} (t)

. Use the following rule:

Using this rule, you get

\begin{array}{c} f^{'} (t) & = \\ = - 2.50 \cdot (- 0.012) \cdot e^{- 0.012 t} \\ = 0.03 e^{- 0.012 t} \end{array}

\begin{array}{c} f^{'} (t) & = \\ = - 2.50 \cdot (- 0.012) \cdot e^{- 0.012 t} \\ = 0.03 e^{- 0.012 t} \end{array}

Thus the function of the reaction rate is

f^{'} (t) = 0.03 e^{- 0.012 t}

You want to find the reaction rate when it is $2.00$ mmol/L. As you saw in Exercise 2, that it happens for $t = 134$ , so you insert 134 and get

\begin{array}{c} f^{'} (134) & = 0.03 e^{- 0.012 \cdot 134} \\ \approx 0.006 \end{array}

\begin{array}{c} f^{'} (134) = 0.03 e^{- 0.012 \cdot 134} \approx 0.006 \end{array}

This means that the reaction rate is

0.006

mmol/L per second when the concentration is

2.00

mmol/L.

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In any case where each value of $x$ , multiplied by each value of $y$ , gives a specific constant $k$ , this is known as inverse proportionality . We usually write this as in the rule box below: $y$ is equal to $k$ divided by $x$ . Our two explanations are exactly the same. Look here:

Example 1

A calculation shows that the expressions are the same:

\begin{array}{c} x \cdot y & = k \\ \frac{x \cdot y}{x} & = \frac{k}{x}, \\ y & = \frac{k}{x} \end{array}

Rule

Two values, $x$ and $y$ , are inversely proportional if

y = \frac{k}{x},

where $k$ is a constant.

Below, in Example 2, you can see a case where $k = 1$ . Here are some rules for remembering what happens for different values of $k$ :

When $k > 0$ is positive, the graph slides outward from the first quadrant (with a positive $x$ -axis and $y$ -axis) and away from the origin.
When $k < 0$ is negative, the graph lies in the fourth quadrant (part of the coordinate system with a positive $x$ -axis and a negative $y$ -axis), but turned upside down. The shape of the graph is always the same.

Example 2

This graph shows $y = \frac{1}{x}$ , so $k = 1$ . Since the graph is inversely proportional, it means that all the coordinates on the graph are such that if you take the $x$ -coordinate and multiply them by the $y$ -coordinate, the answer is $k = 1$ .

$\"Example$

Example 3

Is the graph $y = \frac{2}{3 x}$ inversely proportional?

You can find this out with a few modifications:

\begin{array}{c} y & = \frac{2}{3 x} = \frac{2 \cdot 1}{3 \cdot x} = \frac{2}{3} \cdot \frac{1}{x} \\ \approx 0.67 \cdot \frac{1}{x} \approx \frac{0.67}{x} \end{array}

\begin{array}{c} y = \frac{2}{3 x} = \frac{2 \cdot 1}{3 \cdot x} = \frac{2}{3} \cdot \frac{1}{x} \approx 0.67 \cdot \frac{1}{x} \approx \frac{0.67}{x} \end{array}

You have now determined that

k = \frac{2}{3} \approx 0.67

, so the graph is inversely proportional.

Example 4

You are given the following points:

$\"Function$

$\"Function$

Do the points follow an inversely proportional function?

You know that if the points lie on the graph of an inversely proportional function, then you get the same answer when you multiply the $x$ -value by the $y$ -value for all the points:

\begin{array}{c} 1 \cdot 20 & = 20 & 2 \cdot 10 & = 20 & 3 \cdot 7 & = 21 \\ 4 \cdot 5 & = 20 & 5 \cdot 4 & = 20 \end{array}

It’s close, but since one answer is not the same as the others, you do not have inverse proportionality.

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A power function is a special case of a polynomial function. A power function consists of only one term—a polynomial of the form $a x^{n}$ .

Theory

Power and Root Functions

A power function is a function where

f (x)

is given as a number multiplied by an arbitrary power of

x

. The function can be written as

f (x) = a \cdot x^{n}

If $n$ is a fraction, we call the power function a root function, since it can be rewritten using the formula

x^{\frac{m}{n}} = \sqrt[n]{x^{m}}

Below is a brief description of how each function behaves for different values of $n$ .

If $n$ is a even number, then you get a parabola.

$\"Graph$
If $n$ is an odd number, you get graphs that are extended along the entire $y$ -axis.

$\"Graph$
If $n = 0$ , you get a straight line that intersects $y = a$ .
$\"Graph$
If $n < 0$ , you get rational functions.

$\"Graph$
If $n \in ℚ$ ( $n$ is a fraction), you get a root function.

$\"Graph$
If $n$ is in the form $n = \frac{k}{2 m}$ , and if $k$ and $2 m$ have no factors in common, then the graph begins in the origin.

$\"Graph$

Note! Root functions are defined only for positive values of $x$ , since you can only take the even root ( $\sqrt{x}$ , $\sqrt[4]{x}$ , $\sqrt[6]{x}$ , $\dots$ ) of numbers greater than or equal to 0.

Example 1

f (x) = 5 x^{\frac{1}{2}} = 5 \sqrt{x}

is a power function and a root equation

Example 2

f (x) = 2 x^{4}

is a power function and a polynomial function.

Example 3

f (x) = 4 x^{- 2} = \frac{4}{x^{2}}

is a power function and a rational function.

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A logarithmic function is a type of function that is always growing, but very slowly. There are many different logarithmic functions, but the ones you are going to learn in this article are the most common.

The two most common logarithms are the common logarithm, $\log (x)$ or $\lg (x)$ , and the natural logarithm, $\ln (x)$ . The functions $\log (x)$ and $\ln (x)$ work in exactly the same way when it comes to arithmetic, and they follow the same rules. Their graphs are also about the same. The only difference is that $\log (x)$ has 10 as its base— that is— $1 0^{\log x} = x$ , while $\ln (x)$ has $e$ as its base, that is, $e^{\ln x} = x$ . The natural logarithm got its name because it has the “natural” number $e$ as its base.

Theory

Logarithmic Function

The logarithms $\ln (\dots)$ and $\log (\dots)$ are functions, so when you take the logarithm of a number ( $x = a$ ) you will get a $y$ -value.

\begin{array}{c} f (x) & = \ln (x), \\ f (x) & = \log (x), \end{array}

Note! Since the exponential function is always positive, and the logarithmic function is the “opposite” function of the exponential function, it’s not possible to take the logarithm of a number that is negative or zero.

In this figure, you see the graphs of the common logarithm and the natural logarithm. Both intersect the $x$ -axis at $x = 1$ , but $\ln x$ grows a little faster than $\log x$ .

$\"Graph$

Example 1

The magnitude of an earthquake is measured using the Richter scale. This is a logarithmic scale, and the function is given by
$R (E) = \frac{\log E + 1.32}{1.44}$

where $E$ is the energy that is released in kWh, and $R$ is called the Richter magnitude (this is the number you read about in the newspaper when they report the magnitude of the earthquake). The Richter magnitude tells you how strong an earthquake is.
You read in the newspaper one day that an earthquake in Japan released $1.4 \cdot 1 0^{9}$ kWh of energy. You wonder what this corresponds on the Richter scale.

To find the Richter magnitude, you have to insert $1.4 \cdot 1 0^{9}$ for $E$ , and compute the function value:

\begin{array}{c} R & = \frac{\log}{1.44} \\ \approx 7.27 \end{array}

\begin{array}{c} R \end{array}

Thus, the magnitude of the earthquake was equal to

7.27

on the Richter scale, which is a pretty strong earthquake!

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Functions with $x$ in the denominator are called rational functions. These are in fractional form and have asymptotes. Asymptotes are invisible lines in the coordinate system that the graph moves toward, but never meets. Rational functions can look like the graphs in the figure below.

Theory

Rational Function

A rational function is expressed in the following form:

f (x) = \frac{g (x)}{h (x)},

where $g (x)$ and $h (x)$ are polynomials.

$\"Graphs$

You can see the graph of the following functions in the picture above

\begin{array}{c} f (x) = \frac{1}{x} \end{array}

\begin{array}{c} f (x) = \frac{1}{x} \end{array}

Hyperbolas are among the simplest rational functions. Hyperbolas are the relationship between two linear functions, $a x + b$ and $c x + d$ .

Theory

Hyperbolas

The hyperbola is an important rational function. It represents the relationship between two linear functions. The formula for a hyperbola is

f (x) = \frac{a x + b}{c x + d} .

$\"Hyperbola$

A hyperbola has a vertical and a horizontal asymptote (the dotted lines).

Formula

The Asymptotes of a Hyperbola

Vertical asymptotes

are found where the denominator is equal to zero. You can use this formula:

x = - \frac{d}{c}

Horizontal asymptote

is the value the graph goes towards when $x \to \pm \infty$ (read as, when x approaches positive or negative infinity). You can use this formula:

y = \frac{a}{c}

Example 1

Consider the following hyperbolic function:

f (x) = \frac{2 x + 2}{x - 1}

Find the asymptotes and the intersection between the graph and the axes.

The general formula for the hyperbola is

f (x) = \frac{a x + b}{c x + d}

In this exercise, $a = 2$ , $b = 2$ , $c = 1$ and $d = - 1$

You can find the vertical asymptote in this way:

x = - \frac{d}{c} = - \frac{(- 1)}{1} = 1 .

The vertical asymptote is $x = 1$ . Then you find the horizontal asymptote:

y = \frac{a}{c} = \frac{2}{1} = 2

The horizontal asymptote is $y = 2$ . The intersection with the $y$ -axis is found by inputting $x = 0$ , because the $x$ -coordinate is 0 along the entire $y$ -axis. This gives

f (0) = \frac{2 \cdot 0 + 2}{0 - 1} = \frac{2}{- 1} = - 2

The hyperbola intersects the $y$ -axis at $(0, - 2)$ . The intersection with the $x$ -axis is found by putting $f (x) = 0$ , because the $y$ -coordinate is 0 along the entire $x$ -axis. This gives

f (x) = \frac{2 x + 2}{x - 1} = 0

It is sufficient to set the numerator equal to zero, since a fraction is zero as long as the numerator is zero. Then you get

\begin{array}{c} 2 x + 2 & = 0 \\ 2 x & = - 2 \\ x & = - 1 \end{array}

The hyperbola intersects the $x$ -axis at $(- 1, 0)$ .

(x_{1}, y_{1})

and

(x_{2}, y_{2})

, it is sufficient that you find the slope of the line passing through the points.

Theory

Average Rate of Change

Given two points $(x_{1}, y_{1}) = (x_{1}, f (x_{1}))$ and $(x_{2}, y_{2}) = (x_{2}, f (x_{2}))$ , you have

\begin{array}{c} average rate of change & = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

\begin{array}{c} average rate of change & = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

The formula says that you can choose the version of the three fractions you need.

The figure below shows what this looks like graphically.

$\"Average$

In the figure, you see the function $f (x)$ (blue graph). You draw the straight line (red line) between $A$ and $B$ . The point $A$ has the coordinate $(x_{1}, y_{1})$ , where $y_{1} = f (x_{1})$ , and $B$ has the coordinate $(x_{2}, y_{2})$ , where $y_{2} = f (x_{2})$ . The red arrow between $x_{1}$ and $x_{2}$ indicates the change in the $x$ -direction, and the red arrow between $f (x_{1})$ and $f (x_{2})$ indicates the change in the $y$ -direction. Study the drawing carefully and make sure you understand it!

Example 1

You’re going on a trip to a friend’s cabin. You start the trip at 12:00 and arrive at 17:00. It is 220 miles to the cottage. What was the average speed of the car?

You see that $x_{1} = 12$ , $x_{2} = 17$ , and that the change in distance traveled is

220

The change in duration is $17$ . You enter this into the formula for average rate of change to find the average speed $\overset{}{v}$ :

\overset{}{v} = \frac{220}{5} = 44

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A straight line is defined by two points, or by one point and the slope of the line. There is an ingenious formula that allows you to find the formula for a straight line using one point and the slope of the line:

Formula

Point-Slope Equation

The formula that defines the line with slope $a$ through the point $(x_{1}, y_{1})$ is

y - y_{1} = a (x - x_{1})

Solve the equation with respect to $y$ and the expression looks like the function for a straight line $y = a x + b$ .

Example 1

Find the function for the line through $(2, 5)$ with a slope of 3

Put the numbers into the point-slope equation and solve for $y$ : $\begin{array}{c} y - 5 & = 3 (x - 2) \\ y & = 3 x - 6 + 5 = 3 x - 1 \end{array}$

Example 2

Find the function for the straight line through the points $(- 3, 9)$ and $(3, - 9)$

First, you find the slope:

a = \frac{- 9 - 9}{3 - (- 3)} = - \frac{18}{6} = - 3

Then, select one of the points in the exercise and put it together with the slope into the point-slope equation:

\begin{array}{c} y - (- 9) & = - 3 (x - 3) \\ y + 9 & = - 3 x + 9 \\ y & = - 3 x \end{array}

Since $b = 0$ , you know that the line passes through the origin. The function for the straight line is

y = - 3 x

If you know the function $f (x)$ , you can use the point-slope equation to find the equation of the tangent line at a point on the graph of $f (x)$ . This is because the slope of the tangent is equal to the value of the derivative of the function $f (x)$ at the same point.

Formula

The Equation for an Arbitrary Tangent

y - y_{1} = f^{'} (x_{1}) (x - x_{1}),

where $(x_{1}, y_{1})$ is a point on the tangent (often the point of tangency) and $f^{'} (x_{1})$ is the slope of the point. When using the formula, you must always solve for $y$ — that is, get $y$ alone on one side.

Example 3

Given the function $f (x) = x^{2} + 3 x - 2$ , find the equation for the tangent at $x = 3$

To fill in the equation, you need values for $y_{1}$ and $f^{'} (x_{1})$ . You know that $x_{1} = 3$ , so $y_{1} = f (3)$ and $f^{'} (x_{1}) = f^{'} (3)$ . We begin by computing $f^{'} (x)$ : $\begin{array}{c} f^{'} (x) & = 2 x + 3 \\ f^{'} (x_{1}) & = f^{'} (3) = 2 (3) + 3 = 9 \\ y_{1} & = f (x_{1}) = f (3) \\ = {(3)}^{2} + 3 (3) - 2 = 16 \end{array}$

Now you put this into the equation and get

\begin{array}{c} y - 16 & = 9 (x - 3) \\ y & = 9 x - 27 + 16 = 9 x - 11 \end{array}

Example 4

Let $g (x) = e^{\sin x}$ . Find the equation for the tangent at $x = 0$ .

You need the values of $f^{'} (x_{1})$ and $y_{1}$ . First, differentiate the function:

f^{'} (x) = \cos x \cdot e^{\sin x}

Now you can calculate $f^{'} (x_{1})$ . Since $x_{1} = 0$ , you get

\begin{array}{c} f^{'} (x_{1}) & = f^{'} (0) \\ = \cos 0 \cdot e^{\sin 0} = 1 \cdot e^{0} = 1 \end{array}

You find $y_{1}$ by putting $x_{1}$ into $f (x)$ :

y_{1} = f (x_{1}) = f (0) = e^{\sin 0} = e^{0} = 1

You put all this into the equation and get

\begin{array}{c} y - 1 & = 1 (x - 0) \\ y & = x + 1 \end{array}

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The instantaneous rate of change is the rate of change at a particular point $P$ . You can find this rate of change by calculating the slope of the tangent at this point $P$ . A tangent is a straight line that touches the graph at that exact point.

Theory

Instantaneous Rate of Change

The instantaneous rate of change at a point $P =$ on the graph of the function $f (x)$ is how much the graph is changing at this particular point. Let the tangent line to $f (x)$ at $x = x_{1}$ be given by

y = a x + b

Then, the instantaneous rate of change at this point is equal to the slope of the tangent. Therefore,

Instantaneous rate of change = a

For a function $f (x)$ and a point $P$ on the graph, you can choose two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on the tangent at $P$ . Then

\begin{matrix} instantaneous rate of change \\ = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{matrix}

\begin{array}{c} instantaneous rate of change & = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array}

If $(x_{1}, y_{1}) = P$ , then

\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{y_{2} - f (x_{1})}{x_{2} - x_{1}}

You can think of the difference between average rate of change and instantaneous rate of change as follows:

The average rate of change can be used to measure speed over a period of time. For example, how long it takes to drive a car $10$ . Instantaneous rate of change, on the other hand, is comparable to what the car’s speedometer is showing at a specific point in time. Instantaneous rate of change means the speed you have at that exact moment.

Example 1

If you have the equation for the tangent at the point where you want to find the instantaneous rate of change, then just look at the slope of the line. But if you only have two points on the tangent, then the method presented here will help you find the slope. You will see that we actually calculate the average rate of change to find the slope. In this figure, you see the function

f (x)

(blue curve):

$\"Example$

You draw the tangent (gray line) at the point $x = x_{1}$ (where $x = x_{1}$ means that the $x$ -coordinate of the point has the value $x_{1}$ , thus $x_{1}$ is a number). Note that one of the two points on the tangent line may be a point on the graph (the point of tangency).

You now choose a point you like on the tangent (gray line) and call it $(x_{2}, y_{2})$ . Now, calculate the slope of the tangent to find the instantaneous rate of change.

You can use the expression for the average rate of change to find an approximation of the instantaneous rate of change at $x_{1}$ . Then you select an $x_{2}$ value very close to the $x$ -value you have been given in the exercise and enter both of them into the formula for the average rate of change.

Example 2

Mark Zuckerberg has started a company and has found this function for the company’s revenue (given in millions of dollars):

f (x) = 0.5 x^{2} - 0.5 x + 2,

where $x$ is the number of weeks. Find the instantaneous rate of change after 4 weeks.

To be able to calculate this, you must draw the graph $f (x)$ in a coordinate system and then draw the tangent at $x = 4$ . To draw the graph you need a function table. Choose some $x$ -values and calculate the $y$ -values:


$x$	$y = f (x)$	$(x, y)$

$0$	$f (0) = 2$	$(0, 2)$

$1$	$f (1) = 2$	$(1, 2)$

$2$	$f (2) = 3$	$(2, 3)$

$3$	$f (3) = 5$	$(3, 5)$

$4$	$f (4) = 8$	$(4, 8)$

$5$	$f (5) = 12$	$(5, 12)$

$6$	$f (6) = 17$	$(6, 17)$

$7$	$f (7) = 23$	$(7, 23)$


$x$	$y = f (x) = 0.5 x^{2} - 0.5 x + 2$	$(x, y)$

$0$	$f (0) = 0.5 \cdot {(0)}^{2} - 0.5 \cdot (0) + 2 = 2$	$(0, 2)$

$1$	$f (1) = 0.5 \cdot {(1)}^{2} - 0.5 \cdot (1) + 2 = 2$	$(1, 2)$

$2$	$f (2) = 0.5 \cdot {(2)}^{2} - 0.5 \cdot (2) + 2 = 3$	$(2, 3)$

$3$	$f (3) = 0.5 \cdot {(3)}^{2} - 0.5 \cdot (3) + 2 = 5$	$(3, 5)$

$4$	$f (4) = 0.5 \cdot {(4)}^{2} - 0.5 \cdot (4) + 2 = 8$	$(4, 8)$

$5$	$f (5) = 0.5 \cdot {(5)}^{2} - 0.5 \cdot (5) + 2 = 12$	$(5, 12)$

$6$	$f (6) = 0.5 \cdot {(6)}^{2} - 0.5 \cdot (6) + 2 = 17$	$(6, 17)$

$7$	$f (7) = 0.5 \cdot {(7)}^{2} - 0.5 \cdot (7) + 2 = 23$	$(7, 23)$

You then get this graph (blue curve):

$\"Example$

Draw the tangent so that it touches only the point $(4, f (4))$ . To find the instantaneous rate of change at $x = 4$ , you must find two points on the tangent line and insert into the formula. You see that the points $(x_{1}, y_{1}) = (2, 1)$ and $(x_{2}, y_{2}) = (4, 8)$ are on the tangent.

Then you get that

\begin{matrix} instantaneous rate of change \\ = \frac{8 - 1}{4 - 2} \\ = \frac{7}{2} \\ = 3.5 \end{matrix}

\begin{array}{c} instantaneous rate of change = \frac{8 - 1}{4 - 2} = \frac{7}{2} = 3.5 \end{array}

You have to multiply this by $

1

to get the correct answer since the revenue is in millions of dollars:

3.5 \cdot $ 1

The instantaneous rate of change of the company’s revenue after four weeks is $ $3$ per week.

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You can approximate the instantaneous rate of change at a point $P$ using average rate of change. To do this, select two points that are very close to each other and near $P$ , so that the slope of the line between them is nearly the slope of the tangent at $P$ .

$\"Approximation$

Approximation is performed because it is difficult to position the tangent accurately. It can often be too slack or too steep, and then the numbers become wrong. When the tangent is too slack, you get a lower value than the answer, and when the tangent is too steep you get a value too high for the answer you are looking for. There is another, more accurate way to calculate the instantaneous growth compared to trying to work with the tangent line. By using the formula for average rate of change and choosing

x_{2}

very close to the point where you will find the growth,

x_{1}

, you’ll get an approximate value for the instantaneous rate of change. Here is an example.

Example 1

Look at $f (x) = 0.5 x^{2} - 0.5 x + 2$ . You want to find the instantaneous rate of change when $x = 4$ by approximation

You use the average rate of change formula:

average rate of change

\begin{array}{c} = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

\begin{array}{c} average rate of change & = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

where

x_{1} = 4

. You’ll now choose

x_{2} = 4.01

(or another value that is very close to

x_{1}

). You then have to find

y_{1} = f (x_{1})

and

y_{2} = f (x_{2})

, and those are

\begin{array}{c} y_{1} & = f (4) \\ = 0.5 {(4)}^{2} - 0.5 (4) + 2 \\ = 8 \\ y_{2} & = f (4.01) \\ = 0.5 {(4.01)}^{2} - 0.5 (4.01) + 2 \\ = 8.035 \end{array}

You enter the values into the formula and get:

\begin{array}{c} approximation & \approx \frac{change in y}{change in x} \\ \approx \frac{8 - 8.035}{4.01 - 4} \\ \approx 3.505 \end{array}

The exact instantaneous rate of change is $3.5$ . You can see by how close the answer here is, that approximation is a pretty good approach.

You can get an even better result if you choose an $x_{2}$ that is even closer to $x_{1}$ . For example, if you select $x_{2} = 4.0001$ , then the calculation works out like this:

\begin{array}{c} y_{2} & = f (4.0001) \\ = 0.5 {(4.0001)}^{2} - 0.5 (4.0001) + 2 \\ = 8.000 \end{array}

You then enter the values into the formula and get:

\begin{array}{c} approximation & = \frac{change in y}{change in x} \\ = \frac{8 - 8.000}{4.0001 - 4} \\ = 3.500 \end{array}

So as you can see, the closer the $x$ -values are to each other, the better the approximation will be. As we saw before, the exact value for instantaneous growth for $x = 4$ is $3.5$ , so this method works well if you use two $x$ values that are very close to each other.

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Limits are values that say something about what happens to an expression when a variable approaches a certain value. The value of the limit can be $- \infty$ , $\infty$ , or any number on the number line.

We use a specific notation for limits:

\lim_{x \to a} expression

You read this as “the limit of expression as $x$ approaches $a$ ”, or “the limit of expression as $x$ tends to $a$ ”.

Here, “ $\lim$ ” is an abbreviation for “limit”. Mathematicians try to make things as simple and intuitive as possible.

Rule

Limits of Polynomials

For the polynomial

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

the limit

\lim_{x \to a} f (x)

is equal to the function value $f (a)$ for all $a \in ℝ$ . Therefore,

\lim_{x \to a} f (x) = f (a)

Example 1

Find the limit of $f (x) = x^{2} + 3 x - 4$ when $x \to 3$

\begin{array}{c} \lim_{x \to 3} f (x) & = \lim_{x \to 3} x^{2} + 3 x - 4 \\ = 3^{2} + 3 \cdot 3 - 4 \\ = 9 + 9 - 4 \\ = 14 \end{array}

Let $f (x)$ be a polynomial of degree $n$ . Thus

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0}

x

approaches plus or minus infinity, then

a_{n} x^{n}

eventually becomes the dominant term in the expression, and it will determine whether the function value of

f (x)

is positive or negative.

Consider the limit of $f (x)$ when $x \to \infty$ . The dominant term affects the sign of the function value of $f (x)$ , since $x^{n}$ tends to positive infinity if $n$ is an even number, and it tends to negative infinity if $n$ is an odd number. Below is a summary of the different cases:

Rule

Limits of Polynomials When $x$ Tends to Plus/Minus Infinity

For $a_{n} > 0$ and $n$ an even number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = \infty \\ \lim_{x \to - \infty} f (x) = \infty \end{matrix}

\lim_{x \to \infty} f (x) = \infty

For

a_{n} < 0

and

n

an even number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = - \infty \\ \lim_{x \to - \infty} f (x) = - \infty \end{matrix}

\lim_{x \to \infty} f (x) = - \infty

For

a_{n} > 0

and

n

an odd number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = \infty \\ \lim_{x \to - \infty} f (x) = - \infty \end{matrix}

\lim_{x \to \infty} f (x) = \infty

For

a_{n} < 0

and

n

an odd number, you’ve got

\begin{matrix} \lim_{x \to \infty} f (x) = - \infty \\ \lim_{x \to - \infty} f (x) = \infty \end{matrix}

\lim_{x \to \infty} f (x) = - \infty

In any case, a polynomial tends to either plus or minus infinity when

x \to \pm \infty

Example 2

Find the limit of $\lim_{x \to - \infty} - 2 x^{3} - 2 x + 5$

Here $f (x)$ is a polynomial of degree 3 and thus of odd degree. In addition, the coefficient of the highest degree term is negative. The limit is therefore

\lim_{x \to - \infty} - 2 x^{3} - 2 x + 5 = \infty

Rule

Trick for Finding Limits of Rational Functions

If both the numerator and the denominator in a fraction tend to zero when $x \to a$ , you can factorize the numerator and denominator separately to find the limit of the fraction $\lim_{x \to a} \frac{f (x)}{g (x)}$ .
If both the numerator and the denominator in a fraction tend to infinity when $x \to a$ , you can divide all the terms in the expression by the highest power of $x$ in the expression.

Note! When you have a simple fraction $\frac{1}{x}$ and $x \to \infty$ , then $\frac{1}{x} = 0$ . All fractions where $x$ is only found in the denominator approach 0 when $x$ approaches infinity. Neat!

Rule

Some Important Limits

When $a \in ℝ ∖$ , the following are true:

\begin{matrix} \lim_{x \to \infty} \frac{a}{x} = 0 \\ \lim_{x \to - \infty} \frac{a}{x} = 0 \\ \lim_{x \to 0} \frac{a}{x} = \infty \end{matrix}

\lim_{x \to \infty} \frac{a}{x} = 0

Note! $\infty$ is not a number! So no matter how big a number you choose on the number line, $\infty$ is infinitely larger. This means that the ratio between $x$ and $a$ when $x \to \infty$ or $x \to - \infty$ is always infinitely large.

Example 3

Find the limit of $\lim_{x \to 1} \frac{x^{2} - x}{x - 1}$

You then have

\lim_{x \to 1} \frac{x^{2} - x}{x - 1} = \lim_{x \to 1} \frac{x (x-1)}{x-1} = \lim_{x \to 1} x = 1

\lim_{x \to 1} \frac{x^{2} - x}{x - 1} = \lim_{x \to 1} \frac{x (x-1)}{x-1} = \lim_{x \to 1} x = 1

Example 4

Find the limit of $\lim_{x \to \infty} \frac{2 x^{2} - 3 x}{x^{2} + 2}$

You then have

\begin{array}{c} \lim_{x \to \infty} \frac{2 x^{2} - 3 x}{x^{2} + 2} & = \lim_{x \to \infty} \frac{}{} \\ = \lim_{x \to \infty} \frac{}{} \\ = \frac{2 - 0}{1 + 0} \\ = 2 \end{array}

Example 5

Find the limit of $\lim_{x \to 0} \frac{x^{2} - 2 x}{x}$

You then have

\begin{array}{c} \lim_{x \to 0} \frac{x^{2} - 2 x}{x} & = \lim_{x \to 0} \frac{x (x - 2)}{x} \\ = \lim_{x \to 0} \frac{x (x - 2)}{x} \\ = \lim_{x \to 0} x - 2 \\ = - 2 \end{array}

Rule

Limits of Exponential Functions

For $0 < a < 1$ you’ve got

\lim_{x \to \infty} a^{x} = 0 and \lim_{x \to - \infty} a^{x} = \infty

For $a > 1$ you’ve got

\lim_{x \to \infty} a^{x} = \infty and \lim_{x \to - \infty} a^{x} = 0

For $a = 1$ you’ve got

\lim_{x \to \infty} 1^{x} = \lim_{x \to - \infty} 1^{x} = 1

Example 6

Look at the function $f (x) = e^{0.5 x}$ . Is the function value approaching zero or infinity when $x \to \infty$ ?

Here, you must rewrite the function to find the growth factor, since it’s not obvious whether the growth factor is greater or less than 1:

e^{0.5 x} = {(e^{0.5})}^{x} \approx 1.64 8^{x}

Thus, you see that

\lim_{x \to \infty} f (x) = \infty

Note! You could have seen that $e^{0.5} > 1$ since $e^{x}$ is a strictly increasing function and $e^{0} = 1$ .

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x

in an interval, and that these points are adjacent to each other, meaning there are no gaps between any of them.

When a graph is continuous, it means that you can draw it without lifting your pencil. By contrast, there are graphs where the $y$ -value of a given $x$ -value is some distance away from the $y$ -value of an adjacent $x$ -value. These graphs are not continuous. Functions that are not defined for all $x$ -values on an interval are also not continuous on that interval. Below, you see drawings of the different cases:

$\"Continuous$

Theory

Continuity

If $\lim_{x \to a} f (x) = f (a)$ , then $f$ is continuous for $x = a$ .
If $\lim_{x \to a} f (x) \neq f (a)$ , then $f$ is discontinuous for $x = a$ .

When

f (x)

is continuous for all

x

in an interval, we say that it is continuous on the interval.

Example 1

Determine whether the function $f (x) = \frac{x^{2} - 2 x}{x - 1}$ is continuous for $x = 1$

\begin{array}{c} \lim_{x \to 1} \frac{x^{2} - 2 x}{x - 1} & = \lim_{x \to 1} \frac{}{} \\ = \lim_{x \to 1} \frac{}{} \\ = \lim_{x \to 1} \frac{1 - \frac{2}{1}}{\frac{1}{1} - \frac{1}{1^{2}}} \\ = \frac{1 - 2}{1 - 1} \\ = not defined \end{array}

You see that

f (x)

is discontinuous, since you got an invalid expression when calculating the limit.

Example 2

Determine if the function $f (x) = \frac{x^{2} - 2 x + 2}{x + 3}$ is continuous for $x = - 2$

\begin{matrix} \lim_{x \to - 2} \frac{x^{2} - 2 x + 2}{x + 3} \\ = \frac{{(- 2)}^{2} - 2 (- 2) + 2}{- 2 + 3} \\ = \frac{4 + 4 + 2}{1} \\ = 10 \end{matrix}

You see that $f (x)$ is continuous since you got a numeric value as the result when you calculated the limit.

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Asymptotes are lines that the graph approaches, but never meets. Asymptotes are imaginary lines that you want to draw as dashed lines, so it’s easy to see where they are while indicating that they aren’t part of the graph.

Theory

Vertical Asymptotes

The line $x = a$ is a vertical asymptote if $f (x) \to \pm \infty$ when $x \to a$ . Vertical asymptotes occur when the denominator of a fraction is zero, because the function is undefined there.

Theory

Horizontal asymptotes

The line

y = c

is a horizontal asymptote if

f (x) \to c

when

x \to \pm \infty

Note! There are also oblique (slanted) asymptotes, but they are not as common.

Example 1

You have the expression $f (x) = \frac{x^{2} + 2 x + 1}{x - 2}$ . Find any vertical and horizontal asymptotes.

Vertical Asymptotes

To find the vertical asymptotes, you need to set the denominator equal to zero:

x - 2 = 0 \Leftrightarrow x = 2

Horizontal Asymptotes

You find the horizontal asymptotes by calculating the limit:

\begin{array}{c} \lim_{x \to \infty} \frac{x^{2} + 2 x + 1}{x - 2} & = \lim_{x \to \infty} \frac{}{} \\ = \lim_{x \to \infty} \frac{}{} \\ = \frac{1 + 0 + 0}{0} \\ \Rightarrow divergent \end{array}

Note! The word “divergent” in this context means that the limit does not exist.

The figure shows the graph of the function in Example 1. As you can see, there is a vertical asymptote at $x = 2$ and no horizontal asymptote. There is an oblique asymptote.

$\"Example$

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The derivative of a function $f (x)$ at $x = x_{1}$ is a number that tells us how much the graph of the function $f (x)$ is increasing (or decreasing) when the $x$ -coordinate is $x_{1}$ . The derivative at $x = x_{1}$ is written as $f^{'} (x_{1})$ . The relationships between the derivative, the instantaneous rate of change, and the slope of the tangent to the graph $f (x)$ is described below.

Theory

The Derivative, the Instantaneous Rate of Change, and the Slope of the Tangent

\begin{array}{c} f^{'} (x_{1}) \\ = the instantaneous rate of \\ change at x_{1} \\ = the slope of the tangent to \\ f (x) in the point \end{array}

\begin{array}{c} f^{'} (x_{1}) & = the instantaneous rate of change at x_{1} \\ = the slope of the tangent to f (x) in the point . \end{array}

Before you read the explanation below, you need to know what $Δ x$ and $x + Δ x$ means ( $Δ x$ is read as “delta $x$ ”).

$Δ x$ means “change in the value of $x$ ”, so it represents the distance between two $x$ -values.
$x + Δ x$ means a distance $Δ x$ from $x$ .

Explanation of the Derivative

Look closely at the figures below while you read the text.

$\"The$

Figure 1: The graph of the function in blue and the secant in pink. The points of intersection are

(x, f (x))

(x + Δ x, f (x + Δ x))

You have the function $f (x)$ (the blue graph), and you have drawn a secant (the pink line) between the points $(x, f (x))$ and $(x + Δ x, f (x + Δ x))$ . As you’ve now learned, the slope of the pink line is the average rate of change of the function between these points. In addition you’ve learned that if these points lie close together, then the average rate of change approaches the instantaneous rate of change.

$\"The$

Figure 2: The same situation as in the previous figure. However, now the point

(x + Δ x, f (x + Δ x))

is closer to

(x, f (x))

, and thus

Δ x

is smaller.

By reducing $Δ x$ , the rightmost point of intersection $(x + Δ x, f (x + Δ x))$ approaches the leftmost point of intersection $(x, f (x))$ . When this happens, the distance between $x$ and $x + Δ x$ , as well as the distance between $f (x)$ and $f (x + Δ x)$ , becomes smaller. The slope of the secant is gradually approaching the slope of the tangent of the function at $(x, f (x))$ .

$\"The$

Figure 3: The same situation as in the two previous figures. However, this time you’ve let

Δ x

approach zero, and the pink line is now tangent to

f (x)

As the distance between the two points of intersection approaches zero, the pink line will touch

f (x)

(x, f (x))

. The slope of the tangent is equal to the instantaneous rate of change at that point.

The value of $x$ was arbitrary. This means that if you formulate your approach mathematically, you get a new function for the slope of $f (x)$ for all values of $x$ . This is precisely what the derivative $f^{'} (x)$ of a function is.

By using what you’ve learned regarding instantaneous rate of change and limits, you get the following definition of the derivative:

Theory

Definition of the Derivative

f^{'} (x) = \lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}

Example 1

Given $f (x) = x^{2}$ , you can differentiate $f (x)$ by using the definition:

\begin{array}{c} f^{'} (x) & = \lim_{Δ x \to 0} \frac{{(x + Δ x)}^{2} - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{(x + Δ x) (x + Δ x) - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{x^{2} + 2 x Δ x + {(Δ x)}^{2} - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{2 x Δ x + {(Δ x)}^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{Δ x (2 x + Δ x)}{Δ x} \\ = \lim_{Δ x \to 0} \frac{Δx (2 x + Δ x)}{Δx} \\ = \lim_{Δ x \to 0} (2 x + Δ x) \\ = 2 x + 0 \\ = 2 x \end{array}

It’s pretty rare to differentiate a function by using the definition of the derivative. Most of the time, you use the rules for derivation.

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Along with continuity, you can also talk about whether or not a function is differentiable. A function is differentiable at a point when it is both continuous at the point and doesn’t have a “cusp”. A cusp shows up if the slope of the function suddenly changes. An example of this can be seen in the image below.

Functions with a “cusp” may come up when you have what is called a piecewise-defined function. That means the function has one expression on one interval, and a different expression on another interval. In the figure below, you can see that $f (x) = x^{2} + 2$ when $x \leq 1$ (the blue graph) and that $f (x) = - 2 x + 5$ when $x > 1$ (the green graph). Mathematically, this is written as

f (x) = {\begin{matrix} x^{2} + 2 & when x ≤ 1, \\ - 2 x + 5 & when x > 1. \end{matrix}

$\"Differentiability$

Even though the graph in this case is continuous at $x = 1$ , it’s not differentiable at $x = 1$ . A cusp occurs where you can draw several tangents to the graph. At points on the graph where you can draw many tangents, the derivative is not defined, and you can say that the function isn’t differentiable.

To explain differentiability properly, you need to know what right and left limits mean.

Theory

Right and Left Limits

A limit of $f (a)$ is a right limit when you approach the point $x = a$ from $x$ -values greater than $a$ . You write
$\lim_{x \to a^{+}} f (x) = f (a)$
A limit of $f (a)$ is a left limit when you approach the point $x = a$ from $x$ -values smaller than $a$ . You write
$\lim_{x \to a^{-}} f (x) = f (a)$

Now let’s take a look at the definition of differentiability.

Theory

Differentiability

You say that a function is differentiable at a point $x = a$ if $f (x)$ is continuous at $x = a$ and

\lim_{x \to a^{-}} f^{'} (x) = \lim_{x \to a^{+}} f^{'} (x)

Note! When you are checking the differentiability of a piecewise-defined function, you use the expression for values less than $a$ in $\lim_{x \to a^{-}} f^{'} (x)$ and the expression for values greater than $a$ in $\lim_{x \to a^{+}} f^{'} (x)$ .

Example 1

Decide whether
$f (x) = {\begin{matrix} x^{2} + 2 & when x ≤ 1, \\ - 2 x + 5 & when x > 1 \end{matrix}$

from the image above is differentiable

To answer this, you first have to check whether $f (x)$ is continuous. Is $\lim_{x \to a} f (x) = f (a)$ for all $a \in ℝ$ ? For $a \leq 1$ , you have

\lim_{x \to a} f (x) = \lim_{x \to a} x^{2} + 2 = a^{2} + 2 = f (a),

and for $a > 1$ , you have

\lim_{x \to a} f (x) = \lim_{x \to a} - 2 x + 5 = - 2 a + 5 = f (a)

\lim_{x \to a} f (x) = \lim_{x \to a} - 2 x + 5 = - 2 a + 5 = f (a)

The limits exist and you can conclude that

f (x)

is continuous.

Now you need to check whether

\lim_{x \to a^{-}} f^{'} (x) = \lim_{x \to a^{+}} f^{'} (x)

for all $x \in ℝ$ . You begin by finding $f^{'} (x)$ :

f^{'} (x) = {\begin{matrix} 2 x & when x ≤ 1 \\ - 2 & when x > 1 \end{matrix}

You check if $f$ is differentiable at the point $x = 1$ . You begin with the left limit:

\lim_{x \to 1^{-}} f^{'} (x) = \lim_{x \to 1^{-}} 2 x = 2 \cdot 1 = 2

Next, the right limit:

\lim_{x \to 1^{+}} f^{'} (x) = \lim_{x \to 1^{+}} - 2 = - 2

Because

\lim_{x \to 1^{-}} f^{'} (x) = 2 \neq - 2 = \lim_{x \to 1^{+}} f^{'} (x),

you know that $f (x)$ is not differentiable at $x = 1$ .

Example 2

Find out where the function
$f (x) = \frac{2 x^{2} - 3}{x^{2} - 4}$

is both continuous and differentiable

As this is a rational function, you know that it is discontinuous where it has vertical asymptotes, which is where its denominator equals 0. That means you have to solve this:

\begin{array}{c} x^{2} - 4 & = 0 \\ x^{2} & = 4 \\ x & = \pm \sqrt{4} = \pm 2 \end{array}

That means the function $f (x)$ is continuous for all values of $x$ except $x = 2$ and $x = - 2$ . You write this mathematically as $x \in ℝ ∖$ (all $x$ in $ℝ$ except $x = - 2$ and $x = 2$ ).

A function has to be continuous at a given point to be differentiable at that point, so you can conclude that the function is not differentiable at the points $x = - 2$ and $x = 2$ . The question is if there are other points where $f (x)$ is not differentiable. You check that by finding out whether

\lim_{x \to a^{-}} f^{'} (x) = \lim_{x \to a^{+}} f^{'} (x)

for all $x$ . First, you find an expression for the derivative:

\begin{array}{c} f^{'} (x) & = \frac{4 x \cdot}{} \\ = \frac{4 x^{3} - 16 x - 4 x^{3} + 6 x}{} \\ = \frac{- 10 x}{} \\ f^{'} (a) & = \frac{- 10 a}{} \end{array}

You know that $f^{'} (a)$ is defined for all $a$ except where the denominator of $f^{'} (a)$ is not defined. This happens when the denominator is 0. You set the denominator equal to 0 and solve for $a$ : $\begin{array}{c} = 0 \\ a^{2} - 4 & = 0 \\ a^{2} & = 4 \\ a & = \pm \sqrt{4} = \pm 2 \end{array}$

That means $f (x)$ is differentiable for all $x \in ℝ ∖$ (all $x$ in $ℝ$ except $x = - 2$ and $x = 2$ ).

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The good news is that you don’t have to differentiate all expressions with the definition of the derivative. Here are all the rules you can follow to be able to differentiate the basic functions.

Look closely at the examples, and make sure you understand what is happening.

Rule

Rules for Derivation

\begin{array}{c} f (x) & = k & \Leftrightarrow & f^{'} (x) & = 0 \\ f (x) & = a x & \Rightarrow & f^{'} (x) & = a \\ f (x) & = x^{n} & \Rightarrow & f^{'} (x) & = n x^{n - 1} \\ f (x) & = \sqrt{x} = x^{\frac{1}{2}} & \Rightarrow & f^{'} (x) & = \frac{1}{2} x^{- \frac{1}{2}} \\ = \frac{1}{2 \sqrt{x}} \\ f (x) & = e^{x} & \Rightarrow & f^{'} (x) & = e^{x} \\ f (x) & = e^{k x} & \Rightarrow & f^{'} (x) & = k e^{k x} \\ f (x) & = a^{x} & \Rightarrow & f^{'} (x) & = a^{x} \ln a \\ f (x) & = \ln x & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \ln (k x) & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \sin x & \Rightarrow & f^{'} (x) & = \cos x \\ f (x) & = \cos x & \Rightarrow & f^{'} (x) & = - \sin x \\ f (x) & = \sin (k x) & \Rightarrow & f^{'} (x) & = k \cos (k x) \\ f (x) & = \cos (k x) & \Rightarrow & f^{'} (x) & = - k \sin (k x) \\ f (x) & = \tan x & \Rightarrow & f^{'} (x) & = \frac{1}{\cos^{2} x} \\ = \tan^{2} x \\ f (x) & = \tan (k x) & \Rightarrow & f^{'} (x) & = \frac{k}{\cos^{2} (k x)} \\ = k + k \tan^{2} (k x) \end{array}

\begin{array}{c} f (x) & = k & \Leftrightarrow & f^{'} (x) & = 0 \\ f (x) & = a x & \Rightarrow & f^{'} (x) & = a \\ f (x) & = x^{n} & \Rightarrow & f^{'} (x) & = n x^{n - 1} \\ f (x) & = \sqrt{x} = x^{\frac{1}{2}} & \Rightarrow & f^{'} (x) & = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}} \\ f (x) & = e^{x} & \Rightarrow & f^{'} (x) & = e^{x} \\ f (x) & = e^{k x} & \Rightarrow & f^{'} (x) & = k e^{k x} \\ f (x) & = a^{x} & \Rightarrow & f^{'} (x) & = a^{x} \ln a \\ f (x) & = \ln x & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \ln (k x) & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \sin x & \Rightarrow & f^{'} (x) & = \cos x \\ f (x) & = \cos x & \Rightarrow & f^{'} (x) & = - \sin x \\ f (x) & = \sin (k x) & \Rightarrow & f^{'} (x) & = k \cos (k x) \\ f (x) & = \cos (k x) & \Rightarrow & f^{'} (x) & = - k \sin (k x) \\ f (x) & = \tan x & \Rightarrow & f^{'} (x) & = \frac{1}{\cos^{2} x} = \tan^{2} x \\ f (x) & = \tan (k x) & \Rightarrow & f^{'} (x) & = \frac{k}{\cos^{2} (k x)} = k + k \tan^{2} (k x) \end{array}

For functions

f (x)

g (x)

and constants

k

, the following applies:

\begin{array}{c} {(f (x) \pm g (x))}^{'} & = f^{'} (x) \pm g^{'} (x) \\ {(k f (x))}^{'} & = k f^{'} (x) \end{array}

In addition to these rules, you can read the entries about the chain rule, the product rule and the quotient rule.

Below are some examples to help you practice these rules.

Example 1

f (x) = 6 \Rightarrow f^{'} (x) = 0

Example 2

f (x) = 12 x \Rightarrow f^{'} (x) = 12

Example 3

f (x) = x^{3} \Rightarrow f^{'} (x) = 3 x^{2}

Example 4

f (x) = 7 x^{5} \Rightarrow f^{'} (x) = 35 x^{4}

Example 5

f (x) = 2 x^{3} - 6 x^{9} \Rightarrow f^{'} (x) = 6 x^{2} - 54 x^{8}

f (x) = 2 x^{3} - 6 x^{9} \Rightarrow f^{'} (x) = 6 x^{2} - 54 x^{8}

Example 6

f (x) = \frac{1}{x} = x^{- 1} \Rightarrow f^{'} (x) = - \frac{1}{x^{2}}

Example 7

f (x) = e^{4 x} \Rightarrow f^{'} (x) = 4 e^{4 x}

Example 8

f (x) = 3^{x} \Rightarrow f^{'} (x) = 3^{x} \ln 3

Example 9

f (x) = \ln 4 x \Rightarrow f^{'} (x) = \frac{1}{x}

Example 10

Jax Teller is driving along a straight country road in Italy. The number $s$ of kilometers Jax has covered after $t$ hours is given by the function $s (t) = 50 t^{2}$ . What can you say about his speed?

You find Jax’s speed after $t$ hours by differentiating the function $s (t)$ :

s^{'} (t) = 50 \cdot 2 t = 100 t

According to the model, Jax increases his speed by $100$ km/h every hour.

After 30 minutes ( $0.5$ hours), his speed is

s^{'} (0.5) = (100 \cdot 0.5) km/h = 50

After 1 hour, his speed is

s^{'} (1) = (100 \cdot 1) km/h = 100

After $2.5$ hours, his speed is

s^{'} (2.5) = (100 \cdot 2.5) km/h = 250

When you drive a long time in a straight direction, you can start experiencing speed blindness, which means you don’t notice that you’re driving too fast!

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The chain rule looks a bit complex in its theoretic form, but it is easy to use as soon as you understand the logic of it. The chain rule is used to differentiate composite functions, which are functions that have another function as an argument.

Formula

The Chain Rule

where $f (g) = f (g (x))$ and $g = g (x)$

The derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

This is how you use the chain rule:

1.: Identify what the inner function and the outer function. To put it simply, the inner function is the one you can cover with a finger, while the outer function is the function that is visible around your finger.
2.: Call the inner function $g (x)$ . When you differentiate, everything becomes a bit clearer if you write just $g$ instead of $g (x)$ .
3.: Differentiate the outer function.
4.: Differentiate the inner function.
5.: Multiply the two functions.
6.: Clean up the expression.

Example 1

Differentiate $f (x) =$

Hold your finger over what is inside the parentheses. You can see that the “power of 5” is on the outside. That means

g = 2 x^{2} + 2 x + 4

is the inner function and ${(g)}^{5}$ is the outer function. That gives you $f (g) = {(g)}^{5}$ and $g = 2 x^{2} + 2 x + 4$ , and

\begin{array}{c} f^{'} (g) & = 5 {(g)}^{4} \\ g^{'} & = 4 x + 2 . \end{array}

Then you get

\begin{array}{c} = 5 {(g)}^{4} \cdot g^{'} \\ = 5 \end{array}

Example 2

Differentiate $f (x) = e^{x^{3} + 2 x}$

You hold your finger over the stuff in the exponent and see that “ $e$ ” is on the outside. That gives you that $g = x^{3} + 2 x$ is the inner function and $e^{g}$ is the outer function. This means that $f (g) = e^{g}$ and $g = x^{3} + 2 x$ , and

\begin{array}{c} f^{'} (g) & = e^{g} \\ g^{'} & = 3 x^{2} + 2 \end{array}

Then you get

\begin{array}{c} = e^{g} \cdot g^{'} \\ = e^{x^{3} + 2 x} \\ = \end{array}

Example 3

Differentiate $f (x) = \ln$

Put your finger over what’s in the parentheses and see that $\ln (g)$ is on the outside. That gives you that $g = 4 x^{2} - 3 x + 2$ is the inner function and $\ln (g)$ is the outer function, and

\begin{array}{c} f^{'} (g) & = \frac{1}{g} \cdot g^{'} \\ g^{'} & = 8 x - 3 \end{array}

Then you get

\begin{array}{c} = \frac{1}{g} \cdot g^{'} \\ = \frac{1}{4 x^{2} - 3 x + 2} \cdot (8 x - 3) \\ = \frac{8 x - 3}{4 x^{2} - 3 x + 2} \end{array}

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The product rule is the rule you use when you have a product of two or more functions.

Let $u (x)$ and $v (x)$ be two functions of $x$ . The product rule tells you how to differentiate the product of these two functions. The product is the new function $u (x) \cdot v (x)$ . To simplify it, you just write $u$ and $v$ for the functions $u (x)$ and $v (x)$ , but it’s important to remember that these are functions of $x$ .

Formula

The Product Rule

{(u v)}^{'} = u^{'} v + u v^{'},

where $u = u (x)$ and $v = v (x)$ .

Example 1

Differentiate the expression $x^{2} \sqrt{2 x}$

Looking at the expression, you decide that $u = x^{2}$ and $v = \sqrt{2 x}$ . That gives you $u^{'} = 2 x$ and $v^{'} = \frac{1}{\sqrt{2 x}}$ , and the differentiation is

\begin{array}{c} = \\ = 2 x \cdot \sqrt{2 x} + x^{2} \cdot \frac{1}{\sqrt{2 x}} \\ = 2 x \sqrt{2 x} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{2 x \sqrt{2 x} \cdot \sqrt{2 x}}{\sqrt{2 x}} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{4 x^{2}}{\sqrt{2 x}} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{5 x^{2}}{\sqrt{2 x}} \end{array}

Example 2

Differentiate the expression $e^{x}$

Here, you decide to say that $u = e^{x}$ and $v = 2 x^{3} + 3 x$ . That gives you $u^{'} = e^{x}$ and $v^{'} = 6 x^{2} + 3$ , and the differentiation becomes

\begin{array}{c} = \\ = e^{x} \cdot \\ = 2 x^{3} e^{x} + 3 x e^{x} + 6 x^{2} e^{x} + 3 e^{x} \\ = e^{x} \end{array}

\begin{array}{c} = \\ = e^{x} \cdot \\ = 2 x^{3} e^{x} + 3 x e^{x} + 6 x^{2} e^{x} + 3 e^{x} \\ = e^{x} \end{array}

Example 3

Differentiate the expression

Here, you decide that $u = x^{2} + x - 1$ and $v = \ln x$ . That gives you $u^{'} = 2 x + 1$ and $v^{'} = \frac{1}{x}$ , and the differentiation becomes

\begin{array}{c} = \\ + \\ = (2 x + 1) \cdot \ln x \\ + \\ = 2 x \ln x + \ln x + x + 1 - \frac{1}{x} \end{array}

\begin{array}{c} = \\ = (2 x + 1) \cdot \ln x + \\ = 2 x \ln x + \ln x + x + 1 - \frac{1}{x} \end{array}

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The quotient rule is the rule that tells you how to take the derivative of a function that is a ratio of two differentiable functions.

Formula

The Quotient Rule

where $u = u (x)$ and $v = v (x)$

Note! Sometimes you get lucky, and it’s possible to simplify the answer. Most of the time that’s not possible though, and you can just leave the fraction as it is.

Example 1

Differentiate the expression $\frac{2 x + 1}{e^{x}}$

Here, $u = 2 x + 1$ and $v = e^{x}$ . You get $u^{'} = 2$ and $v^{'} = e^{x}$ , which gives you

\begin{array}{c} = \frac{{(2 x + 1)}^{'} \cdot e^{x} - (2 x + 1) \cdot}{} \\ = \frac{2 \cdot e^{x} - (2 x + 1) \cdot e^{x}}{} \\ = \frac{2 e^{x} - 2 x e^{x} - e^{x}}{} \\ = \frac{e^{x} - 2 x e^{x}}{} \\ = \frac{e^{x} (1 - 2 x)}{} \\ = \frac{1 - 2 x}{e^{x}} . \end{array}

\begin{array}{c} = \frac{{(2 x + 1)}^{'} \cdot e^{x} - (2 x + 1) \cdot}{} \\ = \frac{2 \cdot e^{x} - (2 x + 1) \cdot e^{x}}{} \\ = \frac{2 e^{x} - 2 x e^{x} - e^{x}}{} \\ = \frac{e^{x} - 2 x e^{x}}{} \\ = \frac{e^{x} (1 - 2 x)}{} \\ = \frac{1 - 2 x}{e^{x}} \end{array}

Example 2

Differentiate the expression $\frac{3 x^{3} - 2 x^{2} + 7}{x - 1}$

Here, you have $u = 3 x^{3} - 2 x^{2} + 7$ and $v = x - 1$ . That means $u^{'} = 9 x^{2} - 2$ and $v^{'} = 1$ , and the derivative is

\begin{array}{c} = \frac{1}{{(x - 1)}^{2}} ( \\ = \frac{}{{(x - 1)}^{2}} \\ = \frac{9 x^{3} - 9 x^{2} - 4 x^{2} + 4 x - 3 x^{3} + 2 x^{2} - 7}{{(x - 1)}^{2}} \\ = \frac{6 x^{3} - 11 x^{2} + 4 x - 7}{{(x - 1)}^{2}} \end{array}

\begin{array}{c} = \frac{}{{(x - 1)}^{2}} \\ = \end{array}

Finding x-values and y-values

Definition of a Function

Definition of a Function

Functional value and argument

Read the Coordinates of a Point

Professional Function Table

Finding x-values and y-values

Domain and Range

Asymptotes, Domain and Range

The Intersection Between Two Graphs

Zeros

The Formula for Linear Functions

Important Properties of the Linear Function

How to Find a Function From Two Points

How to Find a Linear Function

Lease “Smart Young 1”

Lease “Smart Young 2”

Quadratic Function

Parabola (Maximum and Minimum)

Finding the Maximum or Minimum of a Parabola

Polynomials

Exponential Functions

The Graph of an Exponential Function

Exponential Functions

The Graph of an Exponential Function

Conversion from b = ek and from ek = b

Power and Root Functions

Logarithmic Function

Rational Function

Hyperbolas

The Asymptotes of a Hyperbola

Average Rate of Change

Point-Slope Equation

The Equation for an Arbitrary Tangent

Instantaneous Rate of Change

Limits of Polynomials

Limits of Polynomials When x Tends to Plus/Minus Infinity

Trick for Finding Limits of Rational Functions

Some Important Limits

Limits of Exponential Functions

Continuity

Vertical Asymptotes

Horizontal asymptotes

Vertical Asymptotes

Horizontal Asymptotes

The Derivative, the Instantaneous Rate of Change, and the Slope of the Tangent

Explanation of the Derivative

Definition of the Derivative

Right and Left Limits

Differentiability

Rules for Derivation

The Chain Rule

The Product Rule

The Quotient Rule

Finding $x$ -values and $y$ -values

Finding $x$ -values and $y$ -values

Conversion from $b = e^{k}$ and from $e^{k} = b$

Limits of Polynomials When $x$ Tends to Plus/Minus Infinity