Use Sophia to knock out your gen-ed requirements quickly and affordably. Learn more
Become a Member Try for Free
×

f'' and Extreme Values of f

Author: Sophia
PDF Version

what's covered
In this lesson, you will use the second derivative to determine if a maximum or minimum occurs at a critical number. Specifically, this lesson will cover:

Table of Contents

1. The Second Derivative Test

When a function has several critical numbers, and when the second derivative is relatively easy to get, the second derivative test is much more efficient to use than the first derivative test when locating maximum and minimum values. However, there are conditions under which the second derivative doesn’t give enough information, and we need to use information from the first derivative to determine maximum and minimum points of a function.

Consider the graph shown below:

A graph with an x-axis and a y-axis ranging from −6 to 6. The graph has an S-shaped curve. The curve descends from the second quadrant, opens upward in the third quadrant at the marked point at (a, f(a)), and then extends into the first quadrant by crossing the y-axis between 0 and −1. In the first quadrant, the curve opens downward at the marked point at (b, f(b)) before passing through (5, 0) and extending into the fourth quadrant. The point (a, f(a)) corresponds to the coordinates (−2, −1.8), and the point (b, f(b)) corresponds to the coordinates (3, 2.4).

Observe the following at the local extreme points:

The point open parentheses a comma space f open parentheses a close parentheses close parentheses is a local minimum.

  • Since there is a horizontal tangent at x equals a comma we know f apostrophe open parentheses a close parentheses equals 0.
  • Since the graph is concave up at x equals a comma we know that f apostrophe apostrophe open parentheses a close parentheses greater than 0.
The point open parentheses b comma space f open parentheses b close parentheses close parentheses is a local maximum.

  • Since there is a horizontal tangent at x equals b comma we know f apostrophe open parentheses b close parentheses equals 0.
  • Since the graph is concave down at x equals b comma we know that f apostrophe apostrophe open parentheses b close parentheses less than 0.
Based on this graph, there is a connection between the concavity of f open parentheses x close parentheses when there is a horizontal tangent line and the type of local extrema at that point. Formally stated, this is called the second derivative test.

hint
Only critical numbers where f apostrophe open parentheses c close parentheses equals 0 are considered for the second derivative test. If f apostrophe open parentheses c close parentheses is undefined, then f apostrophe apostrophe open parentheses c close parentheses is also undefined, which means we cannot determine if f apostrophe apostrophe open parentheses c close parentheses is positive, negative, or zero.

That said, if f open parentheses x close parentheses has critical numbers where f apostrophe open parentheses c close parentheses is undefined, then the first derivative test will need to be used to determine if any local extrema occur at x equals c.

term to know
Second Derivative Test
Suppose f apostrophe open parentheses c close parentheses equals 0 comma which means f open parentheses x close parentheses has a horizontal tangent at x equals c.
  • If f apostrophe apostrophe open parentheses c close parentheses less than 0 comma this means f open parentheses x close parentheses is concave down around c, which means there is a local maximum at c.
  • If f apostrophe apostrophe open parentheses c close parentheses greater than 0 comma this means f open parentheses x close parentheses is concave up around c, which means there is a local minimum at c.
  • If f apostrophe apostrophe open parentheses c close parentheses equals 0 comma the test is inconclusive, and the first derivative test needs to be used to determine the behavior at c.

2. Locating Maximum/Minimum With the Second Derivative Test

Let’s look at a few examples of how the second derivative test is implemented.

EXAMPLE

Determine the local maximum and minimum values of f open parentheses x close parentheses equals short dash 2 x cubed plus 6 x squared plus 15.

First, find the values of c for which f apostrophe open parentheses c close parentheses equals 0.

f open parentheses x close parentheses equals short dash 2 x cubed plus 6 x squared plus 15 Start with the original function; the domain is all real numbers.
f apostrophe open parentheses x close parentheses equals short dash 6 x squared plus 12 x Take the derivative.
short dash 6 x squared plus 12 x equals 0
short dash 6 x open parentheses x minus 2 close parentheses equals 0
short dash 6 x equals 0 comma space x minus 2 equals 0
x equals 0 comma space x equals 2 		Set f apostrophe open parentheses x close parentheses equals 0 and solve.

The critical numbers are x equals 0 and x equals 2.

Now, take the second derivative using f apostrophe open parentheses x close parentheses equals short dash 6 x squared plus 12 x and substitute x equals 0 and x equals 2.

f apostrophe open parentheses x close parentheses equals short dash 6 x squared plus 12 x Take the first derivative.
f apostrophe apostrophe open parentheses x close parentheses equals short dash 12 x plus 12 Take the second derivative.
f apostrophe apostrophe open parentheses 0 close parentheses equals short dash 12 open parentheses 0 close parentheses plus 12 equals 12 Substitute the critical number, x equals 0. Since f apostrophe apostrophe open parentheses 0 close parentheses is positive, f open parentheses 0 close parentheses is a local minimum.
f apostrophe apostrophe open parentheses 2 close parentheses equals short dash 12 open parentheses 2 close parentheses plus 12 equals short dash 12 Substitute the critical number, x equals 2. Since f apostrophe apostrophe open parentheses 2 close parentheses is negative, f open parentheses 2 close parentheses is a local maximum.

Therefore, the local minimum value is f open parentheses 0 close parentheses equals 15 and the local maximum value is f open parentheses 2 close parentheses equals 23. On the graph, the local minimum is located at (0, 15) and the local maximum is located at (2, 23).

Let’s look at another example.

EXAMPLE

Determine the local maximum and minimum values of f open parentheses x close parentheses equals 5 x plus 20 over x.

First, find the values of c for which f apostrophe open parentheses c close parentheses equals 0.

f open parentheses x close parentheses equals 5 x plus 20 over x equals 5 x plus 20 x to the power of short dash 1 end exponent Start with the original function; rewrite to use the power rule.
The domain is open parentheses short dash infinity comma space 0 close parentheses union open parentheses 0 comma space infinity close parentheses.
f apostrophe open parentheses x close parentheses equals 5 minus 20 x to the power of short dash 2 end exponent Take the derivative.
5 minus 20 x to the power of short dash 2 end exponent equals 0
5 minus 20 over x squared equals 0
5 x squared minus 20 equals 0
5 open parentheses x squared minus 4 close parentheses equals 0
5 open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses equals 0
x equals 2 comma space x equals short dash 2 		Set f apostrophe open parentheses x close parentheses equals 0 and solve.

The critical numbers are x equals short dash 2 and x equals 2.

Now, take the second derivative using f apostrophe open parentheses x close parentheses equals 5 minus 20 x to the power of short dash 2 end exponent comma then substitute x equals short dash 2 and x equals 2.

f apostrophe open parentheses x close parentheses equals 5 minus 20 x to the power of short dash 2 end exponent Take the first derivative.
f apostrophe apostrophe open parentheses x close parentheses equals 40 x to the power of short dash 3 end exponent equals 40 over x cubed Take the second derivative.
f apostrophe apostrophe open parentheses short dash 2 close parentheses equals 40 over open parentheses short dash 2 close parentheses cubed equals short dash 5 Substitute the critical number, x equals short dash 2. Since f apostrophe apostrophe open parentheses short dash 2 close parentheses is negative, f open parentheses short dash 2 close parentheses is a local maximum.
f apostrophe apostrophe open parentheses 2 close parentheses equals 40 over 2 cubed equals 5 Substitute the critical number, x equals 2. Since f apostrophe apostrophe open parentheses 2 close parentheses is positive, f open parentheses 2 close parentheses is a local minimum.

Therefore, the local maximum value is f open parentheses short dash 2 close parentheses equals short dash 20 and the local minimum value is f open parentheses 2 close parentheses equals 20. On the graph, the local maximum is located at (-2, -20) and the local minimum is located at (2, 20).

Here is a graph of the function:

A graph with an x-axis and a y-axis ranging from −40 to 40 in increments of 10. The graph has two narrow curves. One curve descends from the first quadrant along the positive y-axis, opens upward from the marked point at (2, 20), and then extends upward very gradually into the same quadrant. The other curve rises gradually from the third quadrant, opens downward at the marked point (−2, −20), and then extends downward into the same quadrant along the negative y-axis.

Let’s now look at an example where the second derivative test cannot be used.

EXAMPLE

Determine the local maximum and minimum values of f open parentheses x close parentheses equals x cubed minus 6 x squared plus 12 x plus 10.

First, find the values of c for which f apostrophe open parentheses c close parentheses equals 0.

f open parentheses x close parentheses equals x cubed minus 6 x squared plus 12 x plus 10 Start with the original function; the domain is all real numbers.
f apostrophe open parentheses x close parentheses equals 3 x squared minus 12 x plus 12 Take the derivative.
3 x squared minus 12 x plus 12 equals 0
3 open parentheses x squared minus 4 x plus 4 close parentheses equals 0
3 open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses equals 0
x equals 2 		Set f apostrophe open parentheses x close parentheses equals 0 and solve.

There is only one critical number, x equals 2.

Now, take the second derivative using f apostrophe open parentheses x close parentheses equals 3 x squared minus 12 x plus 12 comma then substitute x equals 2.

f apostrophe open parentheses x close parentheses equals 3 x squared minus 12 x plus 12 Take the first derivative.
f apostrophe apostrophe open parentheses x close parentheses equals 6 x minus 12 Take the second derivative.
f apostrophe apostrophe open parentheses 2 close parentheses equals 6 open parentheses 2 close parentheses minus 12 equals 0 Substitute the critical number, x equals 2.

This means that the second derivative cannot be used to determine if f open parentheses x close parentheses attains a local maximum or minimum at x equals 2.

To determine the behavior of f open parentheses x close parentheses at x equals 2 comma we now turn back to the first derivative, f apostrophe open parentheses x close parentheses equals 3 x squared minus 12 x plus 12.

Evaluate on either side by finding f apostrophe open parentheses 1 close parentheses and f apostrophe open parentheses 3 close parentheses colon
  • f apostrophe open parentheses 1 close parentheses equals 3 open parentheses 1 close parentheses squared minus 12 open parentheses 1 close parentheses plus 12 equals 3 (increasing)
  • f apostrophe open parentheses 3 close parentheses equals 3 open parentheses 3 close parentheses squared minus 12 open parentheses 3 close parentheses plus 12 equals 3 (increasing)
Since f open parentheses x close parentheses is increasing on both sides of x equals 2 comma this means that there is no minimum or maximum when x equals 2.

The graph of f open parentheses x close parentheses is shown in the figure:

A graph with an x-axis and a y-axis ranging from −40 to 40 in increments of 10. The graph has a curve that rises upward from the third quadrant and passes through the point (1.8, 10) and the marked point at (4, 18) before extending upward into the first quadrant.

watch
In this video, we’ll use the second derivative test to locate local maximum and minimum values of the function f open parentheses x close parentheses equals x e to the power of short dash 0.5 x end exponent.

summary
In this lesson, you learned that under certain conditions, the second derivative can be used to determine if f open parentheses x close parentheses has a local maximum or minimum at a critical value c for which f apostrophe open parentheses c close parentheses equals 0, known as the second derivative test. Next, you explored several examples locating the maximum and minimum values with the second derivative test; however, when those conditions aren’t met, the first derivative test is used to determine the locations of local maximum and minimum values.

Source: THIS TUTORIAL HAS BEEN ADAPTED FROM CHAPTER 3 OF "CONTEMPORARY CALCULUS" BY DALE HOFFMAN. ACCESS FOR FREE AT WWW.CONTEMPORARYCALCULUS.COM. LICENSE: CREATIVE COMMONS ATTRIBUTION 3.0 UNITED STATES.

Terms to Know
Second Derivative Test

Suppose f apostrophe open parentheses c close parentheses equals 0, which means f open parentheses x close parentheses has a horizontal tangent at x equals c.

• If f apostrophe apostrophe open parentheses c close parentheses less than 0, this means f open parentheses x close parentheses is concave down around c, which means there is a local maximum at c.

• If f apostrophe apostrophe open parentheses c close parentheses greater than 0, this means f open parentheses x close parentheses is concave up around c, which means there is a local minimum at c.

• If f apostrophe apostrophe open parentheses c close parentheses equals 0, the test is inconclusive, and the first derivative test needs to be used to determine the behavior at c.

© 2026 SOPHIA Learning, LLC. SOPHIA is a registered trademark of SOPHIA Learning, LLC.