Using the Second Derivative Test to Determine Extrema - AP Calculus AB Study Guide

The Core Idea: Using the Second Derivative Test to Determine Extrema

After identifying the critical points of a function—locations where the first derivative is zero or undefined—the next task is to determine whether these points correspond to relative maxima, relative minima, or neither. The Second Derivative Test provides a powerful and often efficient alternative to the First Derivative Test for this classification.

The core idea is to use the function's concavity at a critical point to determine the nature of the extremum. If a function has a horizontal tangent ($f^{\prime }(c)=0$) and is concave up ($f^{\prime \prime }(c)>0$) at that point, it must be at the bottom of a "valley," indicating a relative minimum. Conversely, if the function has a horizontal tangent and is concave down ($f^{\prime \prime }(c)<0$) at that point, it must be at the top of a "hill," indicating a relative maximum. This test directly links the sign of the second derivative at a critical point to the local behavior of the function.

The Second Derivative Test

The Second Derivative Test is a theorem used to classify a critical point $x=c$ where $f^{\prime }(c)=0$.

Let $f$ be a function such that $f^{\prime }(c)=0$ and the second derivative of $f$, $f^{\prime \prime }$, exists on an open interval containing $c$.

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

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

3. If $f^{\prime \prime }(c)=0$ or $f^{\prime \prime }(c)$ does not exist, the test is inconclusive. Another method, such as the First Derivative Test, must be used to classify the critical point at $x=c$.

Understanding the Conditions and Limitations

It is critical to understand when the Second Derivative Test can and cannot be applied.

Condition 1: The Test Only Applies When $f^{\prime }(c)=0$

The Second Derivative Test is designed exclusively for critical points where the tangent line is horizontal. It cannot be used to classify critical points where the first derivative is undefined (e.g., at a cusp or a corner). For such points, the First Derivative Test is the necessary tool.

Condition 2: The Test Can Be Inconclusive

A result of $f^{\prime \prime }(c)=0$ does not provide any information about the behavior of the function at $x=c$. The point could be a relative maximum, a relative minimum, or neither. For example:

• For $f(x)=x^4$, the critical point is $x=0$. Here, $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$. This function has a relative minimum at $x=0$.

• For $g(x)=-x^4$, the critical point is $x=0$. Here, $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)=0$. This function has a relative maximum at $x=0$.

• For $h(x)=x^3$, the critical point is $x=0$. Here, $h^{\prime }(0)=0$ and $h^{\prime \prime }(0)=0$. This function has neither a maximum nor a minimum at $x=0$.

Because all three outcomes are possible when $f^{\prime \prime }(c)=0$, the test is deemed inconclusive. In this situation, you must revert to the First Derivative Test to analyze the sign of $f^{\prime }(x)$ on either side of $x=c$ to make a conclusion.

Core Concepts & Rules

• The Second Derivative Test is a method for classifying critical points as relative (local) extrema.

• Prerequisite: The test can only be applied at a critical point $c$ where the first derivative is zero ($f^{\prime }(c)=0$).

• Relative Minimum: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)>0$, the function has a relative minimum at $x=c$. The function is concave up at this horizontal tangent.

• Relative Maximum: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0$, the function has a relative maximum at $x=c$. The function is concave down at this horizontal tangent.

• Inconclusive Test: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$ or $f^{\prime \prime }(c)$ is undefined, the Second Derivative Test fails to provide a conclusion. You must use the First Derivative Test to classify the critical point.

Step-by-Step Example 1: Basic Application

Find and classify the relative extrema of the function $f(x)=2x^3+3x^2-12x+1$ using the Second Derivative Test.

Step 1: Find the first derivative.

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

Step 2: Find the critical points where $f^{\prime }(x)=0$.

Set the first derivative equal to zero and solve for $x$.

$$6x^2+6x-12=0$$

$$6(x^2+x-2)=0$$

$$6(x+2)(x-1)=0$$

The critical points are $x=-2$ and $x=1$.

Step 3: Find the second derivative.

$$f^{\prime \prime }(x)=\frac{d}{dx}(6x^2+6x-12)=12x+6$$

Step 4: Apply the Second Derivative Test to each critical point.

• Test $x=-2$:

  Evaluate $f^{\prime \prime }(-2)$:

  $$f^{\prime \prime }(-2)=12(-2)+6=-24+6=-18$$

  Since $f^{\prime \prime }(-2)<0$, the function $f$ has a relative maximum at $x=-2$.

• Test $x=1$:

  Evaluate $f^{\prime \prime }(1)$:

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

  Since $f^{\prime \prime }(1)>0$, the function $f$ has a relative minimum at $x=1$.

Step 5: State the conclusion.

The function $f(x)$ has a relative maximum at $x=-2$ and a relative minimum at $x=1$.

Step-by-Step Example 2: Exam-Style Application

Find and classify the relative extrema of $f(x)=3x^5-5x^3$.

Step 1: Find the first derivative.

$$f^{\prime }(x)=15x^4-15x^2$$

Step 2: Find the critical points where $f^{\prime }(x)=0$.

$$15x^4-15x^2=0$$

$$15x^2(x^2-1)=0$$

$$15x^2(x-1)(x+1)=0$$

The critical points are $x=-1$, $x=0$, and $x=1$.

Step 3: Find the second derivative.

$$f^{\prime \prime }(x)=\frac{d}{dx}(15x^4-15x^2)=60x^3-30x$$

Step 4: Apply the Second Derivative Test to each critical point.

• Test $x=-1$:

  $$f^{\prime \prime }(-1)=60(-1{)}^3-30(-1)=-60+30=-30$$

  Since $f^{\prime \prime }(-1)<0$, $f$ has a relative maximum at $x=-1$.

• Test $x=1$:

  $$f^{\prime \prime }(1)=60(1{)}^3-30(1)=60-30=30$$

  Since $f^{\prime \prime }(1)>0$, $f$ has a relative minimum at $x=1$.

• Test $x=0$:

  $$f^{\prime \prime }(0)=60(0{)}^3-30(0)=0$$

  Since $f^{\prime \prime }(0)=0$, the Second Derivative Test is inconclusive for the critical point at $x=0$.

Step 5: Use the First Derivative Test for the inconclusive point $x=0$.

We must analyze the sign of $f^{\prime }(x)=15x^2(x^2-1)$ in intervals around $x=0$.

• Interval $-1<x<0$: Choose a test value, e.g., $x=-0.5$.

  $f^{\prime }(-0.5)=15(-0.5{)}^2((-0.5{)}^2-1)=15(\text{positive})(\text{negative})=\text{negative}$.

• Interval $0<x<1$: Choose a test value, e.g., $x=0.5$.

  $f^{\prime }(0.5)=15(0.5{)}^2((0.5{)}^2-1)=15(\text{positive})(\text{negative})=\text{negative}$.

Since the sign of $f^{\prime }(x)$ does not change at $x=0$ (it is negative on both sides), there is neither a maximum nor a minimum at $x=0$.

Step 6: State the final conclusion.

The function $f(x)$ has a relative maximum at $x=-1$, a relative minimum at $x=1$, and no extremum at $x=0$.

Using Your Calculator

The Second Derivative Test is an analytical method that must be justified by showing the evaluation of the derivatives. A calculator is best used to verify your results.

To check the classification of a critical point $c$ for a function f(x)`: 1. Enter the function into `Y1`. 2. Use the calculator's derivative functionality to find the second derivative. On a TI-84, you can enter `Y2 = nDeriv(nDeriv(Y1, X, X), X, X)`. *Note: This can be slow and graphically inaccurate, but is useful for evaluation.* 3. After analytically finding a critical point $c where $f^{\prime }(c)=0$, you can check the sign of $f^{\prime \prime }(c)$ on the home screen. For example, to check $f^{\prime \prime }(1)$ for the function in Example 1, you would evaluate Y2(1).

4. The calculator will return a value. The sign of this value ($+$ or -`) should match your analytical work. If the value is very close to zero, it confirms that the test is likely inconclusive. 5. You can also graph `Y1` and use the `CALC` menu's $maximum and $minimum$ features to visually confirm the locations of the extrema you found.

AP Exam Quick Hit

Common Question Types

• Direct Application: Given a function like $f(x)=x{e}^x$, you will be asked to find and classify all relative extrema. You would find $f^{\prime }(x)$ and $f^{\prime \prime }(x)$ and apply the test.

• Inconclusive Case Analysis: You will be given a function, like in Example 2, where $f^{\prime \prime }(c)=0$ for a critical point $c$. The question will test your ability to recognize that the test is inconclusive and then correctly apply the First Derivative Test to classify the point.

• Conceptual Justification: You will be given information and asked to draw a conclusion. For example: "Given that a function $g(x)$ is twice-differentiable, $g^{\prime }(4)=0$, and $g^{\prime \prime }(4)=-5$. What can be concluded about $g(x)$ at $x=4$?" You must conclude that $g(x)$ has a relative maximum at $x=4$ because the first derivative is zero and the second derivative is negative.

Common Mistakes

• Confusing the Conditions: A very common error is to mix up the conclusions: thinking $f^{\prime \prime }(c)>0$ implies a maximum or $f^{\prime \prime }(c)<0$ implies a minimum. Remember: positive second derivative means concave up (a "cup"), which holds a minimum.

• Misapplying the Test: Applying the test to a point where $f^{\prime }(c)$ is undefined or not zero. The test is only valid for critical points with a horizontal tangent.

• Assuming $f^{\prime \prime }(c)=0$ Means No Extremum: Incorrectly concluding that if the test is inconclusive, there is no extremum. You must use the First Derivative Test to make a final determination.

• Forgetting to Test All Critical Points: Finding all critical points but only applying the test to one or two of them.

• Algebraic Errors: Simple mistakes in calculating $f^{\prime }(x)$ or $f^{\prime \prime }(x)$ will lead to incorrect critical points or incorrect conclusions. Always double-check your derivatives.