Calculus 1 Quiz: Second Derivative Test

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Second Derivative TestQuestion 1 of 11

A function $f(x)$ is twice differentiable and has a critical point at $x=2$ . If the second derivative is given by $f''(x) = \ln(x^2 - 2x + 2)$ , what does the Second Derivative Test imply about the point $x=2$ ?

f

has a local minimum at

x=2

f

has a local maximum at

x=2

f

has an inflection point at

x=2

The test is inconclusive.

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Calculus 1 Quiz

Calculus 1 Quiz: Second Derivative Test

Practice Second Derivative Test in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Second Derivative Test, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$f$ has a local minimum at $x=2$ . (correct answer)
$f$ has a local maximum at $x=2$ .
$f$ has an inflection point at $x=2$ .
The test is inconclusive.

Explanation: We are given that

x=2

is a critical point, so we know

f'(2)=0

. To use the Second Derivative Test, we must evaluate the sign of

f''(2)

. Plugging

x=2

into the expression for the second derivative gives:

f''(2) = \ln(2^2 - 2(2) + 2) = \ln(4 - 4 + 2) = \ln(2)

. Since

e^0=1

and

e^1\approx2.718

, we know that

0 < \ln(2) < 1

. Because

f''(2) = \ln(2)

is a positive number, the function is concave up at the critical point

x=2

. Therefore, the Second Derivative Test implies that

f

has a local minimum at

x=2

Question 2

Let $f$ and $g$ be twice-differentiable functions. Suppose $g(1)=2$ , $g'(1)=0$ , and $g''(1)=3$ . Also, $f'(2)=-4$ . Let $h(x) = f(g(x))$ . Use the Second Derivative Test to classify the point $x=1$ for the function $h(x)$ .

$h$ has a local maximum at $x=1$ . (correct answer)
$h$ has a local minimum at $x=1$ .
The test is inconclusive because $g'(1)=0$ .
More information about $f''$ is needed to classify the point.

Explanation: First, determine if

x=1

is a critical point for

h(x)

. Using the chain rule,

h'(x) = f'(g(x))g'(x)

. At

x=1

h'(1) = f'(g(1))g'(1) = f'(2) \cdot 0 = 0

. Since

h'(1)=0

x=1

is a critical point. Next, find the second derivative of

h(x)

using the product rule and chain rule:

h''(x) = f''(g(x))[g'(x)]^2 + f'(g(x))g''(x)

. Now evaluate at

x=1

h''(1) = f''(g(1))[g'(1)]^2 + f'(g(1))g''(1) = f''(2)[0]^2 + f'(2)(3) = 0 + (-4)(3) = -12

. Since

h''(1) = -12 < 0

, the Second Derivative Test indicates that

h(x)

has a local maximum at

x=1

Question 3

Let $f(x)$ be a twice-differentiable function with critical points at $x=2$ and $x=6$ . The graph of its second derivative, $f''(x)$ , is a parabola opening downward with roots at $x=1$ and $x=7$ . Which of the following statements must be true?

$f$ has a local minimum at $x=2$ and a local maximum at $x=6$ .
$f$ has a local maximum at $x=2$ and a local minimum at $x=6$ .
$f$ has local maxima at both $x=2$ and $x=6$ .
$f$ has local minima at both $x=2$ and $x=6$ . (correct answer)

Explanation: The graph of

f''(x)

is a downward-opening parabola with roots at

x=1

and

x=7

. This means

f''(x)

is positive for

x

between 1 and 7, and negative otherwise. We need to classify the critical points

x=2

and

x=6

. Both of these points lie in the interval

(1, 7)

. For any

x

(1, 7)

, the value of

f''(x)

is positive. Therefore,

f''(2) > 0

and

f''(6) > 0

. According to the Second Derivative Test, if

f'(c)=0

and

f''(c)>0

, then

f

has a local minimum at

x=c

. Since this condition holds for both

x=2

and

x=6

, the function

f

must have local minima at both points.

Question 4

A student is asked to classify the critical point of $f(x) = (1-x)e^x$ . Their work is shown below:

$f'(x) = (-1)e^x + (1-x)e^x = (-1 + 1 - x)e^x = -xe^x$ .
Set $f'(x)=0$ , which implies $-xe^x=0$ , so $x=0$ is the only critical point.
$f''(x) = (-1)e^x + (-x)e^x = (-1-x)e^x$ .
Evaluate $f''(0) = (-1-0)e^0 = -1$ .
Conclusion: Since $f''(0) < 0$ , there is a local minimum at $x=0$ .

In which step does the student's first error appear?

Step 1
Step 3
Step 4
Step 5 (correct answer)

Explanation: Let's check each step. Step 1 correctly applies the product rule. Step 2 correctly solves for the critical point. Step 3 correctly finds the second derivative. Step 4 correctly evaluates the second derivative at the critical point. Step 5 correctly identifies that

f''(0) < 0

, but draws the wrong conclusion. The condition

f'(c)=0

and

f''(c)<0

implies that the function is concave down at the critical point, which corresponds to a local maximum, not a local minimum. The first error is in the final conclusion drawn from the results of the test.

Question 5

Consider the function $f(x) = (x-3)^4 + 5$ . When using the Second Derivative Test to analyze the critical point at $x=3$ , what is the conclusion?

The function has a local minimum at $x=3$ .
The function has a local maximum at $x=3$ .
The test is inconclusive for the critical point at $x=3$ . (correct answer)
The function has an inflection point at $x=3$ , not an extremum.

Explanation: The Second Derivative Test requires evaluating the second derivative at a critical point. First, find the critical points by setting the first derivative to zero.

f'(x) = 4(x-3)^3

. Setting

f'(x)=0

gives

x=3

. Next, find the second derivative:

f''(x) = 12(x-3)^2

. Now, evaluate the second derivative at the critical point:

f''(3) = 12(3-3)^2 = 0

. When the second derivative is zero at a critical point, the Second Derivative Test is inconclusive. It does not provide information about whether the point is a maximum, minimum, or neither. While further analysis with the First Derivative Test would show

x=3

is a local minimum, the question specifically asks for the conclusion from the Second Derivative Test itself.

Question 6

The function $f(x) = x - 2\cos(x)$ has a local maximum on the interval $(0, 2\pi)$ at which of the following $x$ -values?

$x = \pi/6$
$x = 5\pi/6$
$x = 7\pi/6$ (correct answer)
$x = 11\pi/6$

Explanation: First, find the critical points by setting

f'(x)=0

. The derivative is

f'(x) = 1 + 2\sin(x)

. Setting this to zero gives

1+2\sin(x)=0

, or

\sin(x) = -1/2

. In the interval

(0, 2\pi)

, the solutions are

x=7\pi/6

and

x=11\pi/6

. Next, use the Second Derivative Test to classify these points. The second derivative is

f''(x) = 2\cos(x)

. Evaluate

f''(x)

at each critical point. At

x=7\pi/6

f''(7\pi/6) = 2\cos(7\pi/6) = 2(-\sqrt{3}/2) = -\sqrt{3} < 0

, which indicates a local maximum. At

x=11\pi/6

f''(11\pi/6) = 2\cos(11\pi/6) = 2(\sqrt{3}/2) = \sqrt{3} > 0

, which indicates a local minimum. Therefore, the local maximum occurs at

x=7\pi/6

Question 7

Suppose $f'(c)=0$ for a twice-differentiable function $f$ . Which of the following conditions is sufficient to conclude that $f$ has a local minimum at $x=c$ ?

$f''(x)$ is positive for all $x$ in an open interval containing $c$ .
$f''(c) \ge 0$ .
$f''(c) = 0$ and $f'''(c) > 0$ .
$f''(c) > 0$ . (correct answer)

Explanation: The Second Derivative Test states that if

f'(c)=0

and

f''(c)>0

, then

f

has a local minimum at

x=c

. Choice D is the precise statement of this test. Choice A is a stronger condition than necessary; we only need to know the sign of

f''(c)

at the point

c

itself. Choice B is incorrect because if

f''(c)=0

, the test is inconclusive. Choice C describes a condition for an inflection point where the function is increasing, not a local minimum (this is related to the Higher-Order Derivative Test, but the conclusion is for an inflection point).

Question 8

Let $f(x)$ be a function such that $f'(3)=0$ and $f''(3)=4$ . If $g(x) = \ln(f(x))$ and $f(3)=e^2$ , what does the Second Derivative Test reveal about the function $g$ at $x=3$ ?

$g$ has a local maximum at $x=3$ .
$g$ has a local minimum at $x=3$ . (correct answer)
The test is inconclusive because $f'(3)=0$ .
The test cannot be applied because $g''(3)$ is undefined.

Explanation: First, we check if

x=3

is a critical point for

g(x)

. Using the chain rule,

g'(x) = \frac{f'(x)}{f(x)}

. At

x=3

g'(3) = \frac{f'(3)}{f(3)} = \frac{0}{e^2} = 0

. So,

x=3

is a critical point. Next, we find the second derivative of

g(x)

using the quotient rule:

g''(x) = \frac{f''(x)f(x) - f'(x)f'(x)}{[f(x)]^2}

. Now we evaluate

g''(3)

g''(3) = \frac{f''(3)f(3) - [f'(3)]^2}{[f(3)]^2} = \frac{(4)(e^2) - (0)^2}{[e^2]^2} = \frac{4e^2}{e^4} = \frac{4}{e^2}

. Since

e>0

e^2>0

, so

g''(3) = 4/e^2 > 0

. Because

g'(3)=0

and

g''(3)>0

, the Second Derivative Test shows that

g(x)

has a local minimum at

x=3

Question 9

Consider the function $f(x) = \frac{x^2 - 8}{x+3}$ . The function has a local maximum at which value of $x$ ?

$x=-4$ (correct answer)
$x=-2$
$x=2$
$x=4$

Explanation: First, find the critical points using the quotient rule for

f'(x)

f'(x) = \frac{2x(x+3) - (x^2-8)(1)}{(x+3)^2} = \frac{2x^2+6x-x^2+8}{(x+3)^2} = \frac{x^2+6x+8}{(x+3)^2} = \frac{(x+4)(x+2)}{(x+3)^2}

. Setting

f'(x)=0

gives critical points at

x=-4

and

x=-2

. (Note

x=-3

is a vertical asymptote). Next, find

f''(x)

. The numerator of

f'(x)

N(x)=x^2+6x+8

and the denominator is

D(x)=(x+3)^2

. Instead of a full quotient rule, it's easier to find the derivative of the simplified

f'(x)

's numerator for the sign of

f''(x)

at critical points, but the formal second derivative is

f''(x) = \frac{(2x+6)(x+3)^2 - (x^2+6x+8) \cdot 2(x+3)}{(x+3)^4} = \frac{(2x+6)(x+3) - 2(x^2+6x+8)}{(x+3)^3} = \frac{2x^2+12x+18 - 2x^2-12x-16}{(x+3)^3} = \frac{2}{(x+3)^3}

. Test the critical points:

f''(-4) = \frac{2}{(-4+3)^3} = -2 < 0

. This indicates a local maximum.

f''(-2) = \frac{2}{(-2+3)^3} = 2 > 0

. This indicates a local minimum. Therefore, the local maximum is at

x=-4

Question 10

Let $f(x) = -x^3 + kx^2 - 10$ for some constant $k$ . If $f$ has a local maximum at $x=2$ , what is the value of $k$ ?

$k = -3$
$k = 3$ (correct answer)
$k = 6$
$k = -6$

Explanation: For

f

to have a local maximum at

x=2

, two conditions must be met:

f'(2)=0

and

f''(2)<0

. First, find the derivative:

f'(x) = -3x^2 + 2kx

. Set

f'(2)=0

-3(2)^2 + 2k(2) = 0 \implies -12 + 4k = 0 \implies k=3

. Now, we must verify this value of

k

satisfies the second condition. The second derivative is

f''(x) = -6x + 2k

. With

k=3

, this becomes

f''(x) = -6x+6

. Evaluating at

x=2

f''(2) = -6(2)+6 = -12+6 = -6

. Since

f''(2) = -6 < 0

, the function is concave down at the critical point, confirming a local maximum. Thus,

k=3

is the correct value.

Question 11

Let $f(x) = \frac{1}{4}x^4 - x^3 - 2x^2 + 5$ . At which of its critical points does $f$ have a local maximum?

$x = -1$
$x = 0$ (correct answer)
$x = 4$
$x = -1$ and $x=4$

Explanation: First, find the critical points by setting the first derivative equal to zero.

f'(x) = x^3 - 3x^2 - 4x = x(x^2 - 3x - 4) = x(x-4)(x+1)

. The critical points are

x=-1

x=0

, and

x=4

. Next, find the second derivative to classify these points:

f''(x) = 3x^2 - 6x - 4

. Now, evaluate

f''(x)

at each critical point: At

x=-1

f''(-1) = 3(-1)^2 - 6(-1) - 4 = 3+6-4 = 5 > 0

. This is a local minimum. At

x=0

f''(0) = 3(0)^2 - 6(0) - 4 = -4 < 0

. This is a local maximum. At

x=4

f''(4) = 3(4)^2 - 6(4) - 4 = 48-24-4 = 20 > 0

. This is a local minimum. Therefore, the function has a local maximum only at

x=0