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

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

The First Derivative Test allows us to classify a critical point as a relative minimum or maximum by analyzing the sign of the first derivative on either side of that point. The Second Derivative Test provides an alternative, and often more efficient, method for this classification. Instead of examining intervals, this test focuses directly on the behavior of the function at the critical point itself.

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

The Second Derivative Test

The Second Derivative Test is a formal method for determining whether a critical point of a function is a relative minimum or a relative maximum.

Let $f$ be a twice-differentiable function on an open interval containing $c$, and suppose that $f^{\prime }(c)=0$.

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 $f^{\prime \prime }(c)=0$, the test is inconclusive. The function $f$ may have a relative minimum, a relative maximum, or neither at $x=c$. Another test (such as the First Derivative Test) must be used.

Understanding the Conditions and the Inconclusive Case

The application of the Second Derivative Test is governed by strict conditions that must be met. The most critical prerequisite is that the test can only be applied to a critical point $c$ where $f^{\prime }(c)=0$. It cannot be used to classify critical points where the first derivative is undefined. Furthermore, the function must be twice-differentiable at $c$, meaning $f^{\prime \prime }(c)$ must exist.

The most important nuance of this test is the inconclusive case, which occurs when $f^{\prime \prime }(c)=0$. It is a common misconception that this result implies there is no extremum at $x=c$. "Inconclusive" means that the Second Derivative Test provides no information about the nature of the critical point.

Consider these two examples:

• For $f(x)=x^4$, we have $f^{\prime }(x)=4x^3$ and $f^{\prime \prime }(x)=12x^2$. At the critical point $x=0$, $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$. The test is inconclusive. However, by analyzing the function, we know $f(x)=x^4$ has a relative minimum at $x=0$.

• For $g(x)=x^3$, we have $g^{\prime }(x)=3x^2$ and $g^{\prime \prime }(x)=6x$. At the critical point $x=0$, $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)=0$. The test is again inconclusive. In this case, $g(x)=x^3$ has neither a relative minimum nor a relative maximum at $x=0$.

Therefore, if you encounter $f^{\prime \prime }(c)=0$, you cannot draw a conclusion from the Second Derivative Test and must revert to the First Derivative Test to classify the critical point at $x=c$.

Core Concepts & Rules

• Purpose: The Second Derivative Test is used to determine if a critical point $c$ where $f^{\prime }(c)=0$ is the location of a relative minimum or a relative maximum.

• Requirement 1: The function $f$ must be twice-differentiable at the point being tested.

• Requirement 2: The test can only be applied at a critical point $c$ where the first derivative is zero ($f^{\prime }(c)=0$).

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

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

• Inconclusive Result: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$, the test fails to provide a conclusion. You must use another method, like the First Derivative Test, to classify the critical point.

Step-by-Step Example 1: Applying the Test to a Polynomial

Find and classify the relative extrema of the function $f(x)=2x^3+3x^2-12x+4$ using the Second Derivative Test.

Step 1: Find the first derivative and identify critical points.

First, we find $f^{\prime }(x)$ and set it equal to zero to find the critical points.

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

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

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

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

The critical points where $f^{\prime }(x)=0$ are $x=-2$ and $x=1$.

Step 2: Find the second derivative.

Next, we compute the second derivative of $f(x)$.

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

Step 3: Apply the Second Derivative Test to each critical point.

We evaluate $f^{\prime \prime }(x)$ at each critical point found in Step 1.

• For $x=-2$:

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

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

• For $x=1$:

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

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

Conclusion:

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

Step-by-Step Example 2: Using Given Derivative Information

A twice-differentiable function $g(x)$ is known to have critical points at $x=-1$ and $x=3$. The second derivative of the function is given by $g^{\prime \prime }(x)=e^x(x^2-2x-2)$. Classify the critical point at $x=3$ as a relative minimum, a relative maximum, or neither. Justify your answer.

Step 1: Identify the conditions for the test.

We are given that $x=3$ is a critical point. For the Second Derivative Test, we assume this means $g^{\prime }(3)=0$. The function is also given as twice-differentiable.

Step 2: Evaluate the second derivative at the critical point.

We need to find the sign of $g^{\prime \prime }(3)$.

$$g^{\prime \prime }(3)=e^3(3^2-2(3)-2)$$

$$g^{\prime \prime }(3)=e^3(9-6-2)$$

$$g^{\prime \prime }(3)=e^3(1)=e^3$$

Step 3: Apply the Second Derivative Test.

We analyze the sign of the result from Step 2.

Since $e^3$ is a positive number, $g^{\prime \prime }(3)>0$.

Step 4: State the conclusion with justification.

Because $g^{\prime }(3)=0$ and $g^{\prime \prime }(3)>0$, the Second Derivative Test guarantees that the function $g(x)$ has a relative minimum at $x=3$.

Using Your Calculator

The Second Derivative Test is an analytical procedure. A calculator cannot perform the test for you, but it can be an extremely powerful tool for performing the necessary calculations, especially with complex functions.

Problem: Use a calculator to help find and classify the relative extrema of $f(x)=x^2\cos (x)$ on the interval $(-\pi ,\pi )$ using the Second Derivative Test.

Steps using a TI-84 style calculator:

1. Define the Derivatives:

   • In the Y= editor, enter Y1 = X^2 * cos(X).

   • Enter Y2 = nDeriv(Y1, X, X). This represents f'(x).

   • Enter Y3 = nDeriv(Y2, X, X). This represents f''(x).

2. Find Critical Points:

   • Graph Y2 (the first derivative) in a suitable window, for example, $Xmin=-3.14$, $Xmax=3.14$.

   • Use the CALC menu (2nd + TRACE) and select 2:zero.

   • Find the x-intercepts of Y2. You will find a zero at $x=0$. By changing the bounds, you will also find zeros near $x\approx -2.289$ and $x\approx 2.289$. Let's call these $c_1,c_2,c_3$.

3. Evaluate the Second Derivative:

   • From the home screen, evaluate Y3 at each critical point found in the previous step.

   • For c_1 \approx -2.289: Type Y3(-2.289) and press ENTER. The result is approximately $9.98$.

   • For $c_2=0$: Type Y3(0) and press ENTER. The result is $2$.

   • For $c_3\approx 2.289$: Type Y3(2.289) and press ENTER. The result is approximately $9.98$.

4. Apply the Test and Conclude:

   • At $x\approx -2.289$, $f^{\prime }(x)\approx 0$ and $f^{\prime \prime }(x)>0$. Therefore, $f$ has a relative minimum.

   • At $x=0$, $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)>0$. Therefore, $f$ has a relative minimum.

   • At $x\approx 2.289$, $f^{\prime }(x)\approx 0$ and $f^{\prime \prime }(x)>0$. Therefore, $f$ has a relative minimum.

Note: The analytical derivatives are $f^{\prime }(x)=2x\cos (x)-x^2\sin (x)$ and $f^{\prime \prime }(x)=(2-x^2)\cos (x)-4x\sin (x)$. The calculator's numerical derivative functions help avoid potential errors in this complex differentiation.

AP Exam Quick Hit

Common Question Types

• Direct Application with Justification: Given a function $f(x)$, you will be asked to find the x-coordinates of its relative extrema and justify your answer using the Second Derivative Test.

  • Example: "Find the x-coordinates of all relative extrema for $f(x)=x-\ln (x^2+1)$. Classify each as a relative maximum or a relative minimum using the Second Derivative Test."

• Abstract Justification from Given Values: You will be given values of the first and second derivatives at a point and asked to draw a conclusion.

  • Example: "Let $h$ be a twice-differentiable function such that $h^{\prime }(-4)=0$ and $h^{\prime \prime }(-4)=5$. Does $h$ have a relative minimum, relative maximum, or neither at $x=-4$? Justify your answer."

• Analysis of the Inconclusive Case: A question designed to see if you correctly understand what happens when $f^{\prime \prime }(c)=0$.

  • Example: "Let $f(x)=(x-2{)}^4$. Show that the Second Derivative Test is inconclusive at the function's only critical point. Then, use another method to classify the critical point."

Common Mistakes

• Forgetting the $f^{\prime }(c)=0$ Prerequisite: The Second Derivative Test only applies at critical points where the first derivative is zero. Students sometimes misapply it to points where $f^{\prime }(c)$ is undefined or to points that are not critical points at all.

• Misinterpreting an Inconclusive Result: A very common error is to state that if $f^{\prime \prime }(c)=0$, then there is no relative extremum at $x=c$. The correct conclusion is that the test provides no information, and another test is required.

• Confusing the Signs: Students often mix up the conclusions. A helpful mnemonic:

  • If $f^{\prime \prime }(c)>0$ (positive), the graph is concave up (like a cup $\cup$), so it has a relative minimum.

  • If $f^{\prime \prime }(c)<0$ (negative), the graph is concave down (like a frown $\cap$), so it has a relative maximum.

• Stating an Incomplete Justification: On the AP Exam, a full justification requires stating both conditions. For example: "Since $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0$, $f$ has a relative maximum at $x=c$." Simply stating that $f^{\prime \prime }(c)<0$ is not sufficient.