House of Math

Make Sign Charts of the Derivatives of a Function

You can use sign charts to analyze the behavior of a function. They help you find maxima, minima and saddle points. Here, it’s important to keep your head in the game. You’re looking to say something about the function $f (x)$ based on its derivative $f^{'} (x)$ . This is how you do it:

Rule

Drawing the Sign Chart of the Derivative

Find the derivative of $f (x)$ . Draw the sign chart of the differentiated function $f^{'} (x)$ . You’re going to find out where this function is above and below the $x$ -axis.
Mark which $x$ -values give you positive $y$ -values (solid line) and which ones give negative $y$ -values (dashed line).

Rule

The Connection Between $f (x)$ and $f^{'} (x)$

It turns out that there is a clear connection between $f (x)$ and $f^{'} (x)$ :

When $f^{'} (x)$ is positive (above the $x$ -axis), $f (x)$ increases.
When $f^{'} (x)$ is negative (below the $x$ -axis), $f (x)$ decreases.
When $f^{'} (x)$ is zero (on the $x$ -axis), $f (x)$ has a maximum, minimum or saddle point.

Rule

Determine Stationary Points

1.: If the zero of $f^{'} (x)$ lies between two solid lines or between two dashed lines, then $f (x)$ has a saddle point—see Figure (a).
2.: If the zero has a solid line on the left and a dashed line on the right, then $f$ increases before the point and decreases after. That means the point is a maximum—see Figure (b).
3.: If the zero has a dashed line on the left and a solid line on the right, then $f$ decreases before the point and increases after. That means the point is a minimum—see Figure (c).

Saddle point of an increasing function

(a)

Maximum of a function

(b)

Minimum of a function

(c)

Example 1

You have a cubic function $f (x) = - \frac{2}{3} x^{3} - x^{2} + 4 x$ . Find the maxima and minima of $f (x)$ .

1.

First, you find the derivative of

f (x)

:

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

2.

You can factorize this expression according to the formula

a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})

where $x_{1}$ and $x_{2}$ are the solutions of $a x^{2} + b x + c = 0$ . That means the factorized expression for the derivative is

\begin{array}{l} f^{'} (x) & = - 2 (x^{2} + x - 2) \\ = - 2 (x + 2) (x - 1) \end{array}

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

3.

Then the sign chart looks like this:

Example of a sign chart of the derivative

4.

Now you have to find out which points are maxima and which are minima. The sign chart tells you that the function decreases until

x = - 2

, increases for a bit until

x = 1

, and then decreases again. That means

x = - 2

is a minimum and

x = 1

is a maximum.

Now you need to find the corresponding $y$ -values. To do so, just insert the $x$ -values you found into the function

f (x) = - \frac{2}{3} x^{3} + x^{2} + 4 x

That gives you:

Minimum point:

\begin{array}{l} (x_{\min}, f (x_{\min})) & = (- 2, f (- 2)) \\ = (2 -, - \frac{20}{3}) \end{array}

because

\begin{array}{l} f (- 2) & = - \frac{2}{3} {(- 2)}^{3} - {(- 2)}^{2} + 4 (- 2) \\ = \frac{16}{3} + 4 - 8 \\ = \frac{16 - 12 - 24}{3} \\ = - \frac{20}{3} \end{array}

\begin{array}{l} f (- 2) & = - \frac{2}{3} {(- 2)}^{3} - {(- 2)}^{2} + 4 (- 2) \\ = \frac{16}{3} + 4 - 8 \\ = \frac{16 - 12 - 24}{3} \\ = - \frac{20}{3} \end{array}

Maximum point:

\begin{array}{l} (x_{\max}, f (x_{\max})) & = (1, f (1)) \\ = (1, \frac{7}{3}) \end{array}

because

\begin{array}{l} f (1) & = - \frac{2}{3} {(1)}^{3} - {(1)}^{2} + 4 (1) \\ = - \frac{2}{3} - 1 + 4 \\ = - \frac{2}{3} - \frac{3}{3} + \frac{12}{3} \\ = \frac{7}{3} . \end{array}

Example of graph illustrated using sign chart

This picture fits with what we found through our calculations. The minimum is at

(- 2, - \frac{20}{3})

and the maximum is at

(1, \frac{7}{3})

. You can see that the graph decreases until it reaches the minimum, because the derivative of the function is negative and its sign chart has a dashed line.

Between the minimum and the maximum the graph increases, because the derivative of the function is positive and its sign chart has a solid line.

From the maximum and onward the graph decreases again, because the derivative of the function is negative and its sign chart has a dashed line.

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