Introduction

Welcome to AP Calculus AB/BC's Topic 2.5 - Using the Second Derivative Test to Determine Extrema! This topic introduces the second derivative test as an alternative method for classifying critical points.

The Second Derivative Test

Technical Definition

Let $c$ be a critical point where $f^{\prime }(c)=0$.

If $f^{\prime \prime }(c)>0$, then $f$ has a local minimum at $x=c$.

If $f^{\prime \prime }(c)<0$, then $f$ has a local maximum at $x=c$.

If $f^{\prime \prime }(c)=0$, then the test is inconclusive and cannot be used to determine an extrema.

Note that $c$ must be a critical point! The test does not work if $c$ is not a critical point.

That's all there is to the second derivative test! If you are confused, it might be helpful to understand why the second derivative test actually works.

Understanding Why The Test Works

In Topic 5.6, we learned that the second derivative can tell us about concavity. Recall that:

1. If $f^{\prime \prime }(c)>0$, then $f(x)$ is concave up at $c$, and the critical point is at the bottom of the function.

2. If $f^{\prime \prime }(c)<0$, then $f(x)$ is concave down at $c$, and the critical point is at the top of the function.

See the graphs below to visual what we just recalled.

Question

1. Use the Second Derivative Test to classify the critical points of
$f(x)=x^3-6x^2+9x$.

Solution

1.
Find the critical points of $f(x)$ by setting $f^{\prime }(x)=0$:

$f(x)=x^3-6x^2+9x$

$f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$

$f^{\prime }(x)=0$

$0=3(x-1)(x-3)$

$x=1$ and $x=3$

Find $f^{\prime \prime }(x)$:

$f^{\prime }(x)=3x^2-12x+9$

$f^{\prime \prime }(x)=6x-12$

Test at $x=1$:

$f^{\prime \prime }(1)=6-12=-6<0$

Because $f^{\prime }(1)=0$ and $f^{\prime \prime }(1)<0$, there must be a local maximum at $x=1$!

Test at $x=3$:

$f^{\prime \prime }(3)=18-12=6>0$

Because $f^{\prime }(3)=0$ and $f^{\prime \prime }(3)>0$, there must be a local minimum at $x=3$!

Notice how, when I justified my answer, I mentioned that $f^{\prime }(c)=0$. This is very important. If $f^{\prime }(c)\ne 0$, then we cannot use the second derivative test.

Comparing First and Second Derivative Tests

Advantages and Disadvantages

Second Derivative Test:

First Derivative Test:

Question

1. Determine the local extrema for $f(x)=2x^3-9x^2+12x-3$ using the second derivative test.

Solution

1.
First, we need to find the critical points for $f$ by setting $f^{\prime }(x)=0$:

$f^{\prime }(x)=6x^2-18x+12=6(x-1)(x-2)$

$f^{\prime }(x)=0$

Therefore the critical points are $x=1$ and $x=2$

Find $f^{\prime \prime }(x)$:

$f^{\prime \prime }(x)=12x-18$

Test at $x=1$:

$f^{\prime \prime }(1)=12-18=-6<0$

Because $f^{\prime }(1)=0$ and $f^{\prime \prime }(1)<0$, there must be a local maximum at $x=1$.

Test at $x=2$:

$f^{\prime \prime }(2)=24-18=6>0$

Because $f^{\prime }(1)=0$ and $f^{\prime \prime }(2)>0$, there must be a local minimum at $x=2$.

When the Second Derivative Test Fails

The Inconclusive Case

When $f^{\prime \prime }(c)=0$, the second derivative test provides no information (this is because zero is neither greater than or less than zero). When this happens, we have to resort to using the first derivative test.

Question

1. Apply the second derivative test to identify local extrema for $f(x)=x^4$.

Find critical points by setting $f^{\prime }(x)=0$:

$f^{\prime }(x)=4x^3$

$f^{\prime }(x)=0$

$0=4x^3$

$x=0$

Find $f^{\prime \prime }(x)$:

$f^{\prime }(x)=4x^3$

$f^{\prime \prime }(x)=12x^2$

Test at $x=0$:

$f^{\prime \prime }(0)=0$

Because $f^{\prime \prime }(0)=0$, the test is inconclusive. As a result, we must use the first derivative test.

We know that for $x<0$, $f^{\prime }(x)<0$. We also know that for $x>0$, $f^{\prime }(x)>0$. Because $f^{\prime }(x)$ changes from negative to positive, we must have a local minimum at $x=0$ via the first derivative test.

Global Extrema from Single Critical Points

If a continuous function has only one critical point on an interval, and that critical point is a local extremum, then it is also the absolute (global) extremum on that interval. This might sound confusing, but with some practice it will make sense.

This theorem saves time, as it means we don't need to check endpoints when there's only one critical point that we've identified as a local extremum.

Question

1. Show that $f(x)=x^2+\frac{4}{x}$ has an absolute minimum on $(0,\infty )$.

Solution

1.
Find critical points by setting $f^{\prime }(x)=0$:

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

$2x-\frac{4}{x^2}=0$

$2x=\frac{4}{x^2}$

$2x^3=4$

$x^3=2$

$x=\sqrt[3]{2}$ 

Notice how this is the only critical point!

Now we can use the second derivative test to determine if $x=\sqrt[3]{2}$ is a local minimum or maximum.

$f^{\prime \prime }(x)=2+\frac{8}{x^3}$

$f^{\prime \prime }\left(\sqrt[3]{2}\right)=2+\frac{8}{2}=6>0$

Because $f^{\prime }(\sqrt[3]{2})=0$ and $f^{\prime \prime }(\sqrt[3]{2})>0$, there is a local minimum at $x=\sqrt[3]{2}$ via the second derivative test.

Because $x=\sqrt[3]{2}$ is the only local minimum on $(0,\infty )$, it must be the absolute minimum on $(0,\infty )$.

Indeed it is, look at the graph for $f(x)$:

Practice Section