Calculus 1 Quiz: First Derivative Test

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First Derivative TestQuestion 1 of 20

The derivative of a function f is given by f'(x) = (x+2)^2(x-1)^3(x-5). At how many distinct x-values does the function f have a local minimum?

Zero

One

Two

Three

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Calculus 1 Quiz

Calculus 1 Quiz: First Derivative Test

Practice First 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 First 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

The derivative of a function f is given by f'(x) = (x+2)^2(x-1)^3(x-5). At how many distinct x-values does the function f have a local minimum?

Zero
One (correct answer)
Two
Three

Explanation: The critical points of f are the roots of f'(x), which are x = -2, x = 1, and x = 5. We test the sign of f'(x) in the intervals defined by these points. The term (x+2)^2 is always non-negative and does not cause a sign change at x = -2. For x < 1, let's test x=0: f'(0) = (+)(-)(-) = positive. For 1 < x < 5, let's test x=2: f'(2) = (+)(+)(-) = negative. For x > 5, let's test x=6: f'(6) = (+)(+)(+) = positive. At x = -2, the sign does not change, so it's not a local extremum. At x = 1, the sign changes from positive to negative, so f has a local maximum. At x = 5, the sign changes from negative to positive, so f has a local minimum. Therefore, f has a local minimum at exactly one x-value.

Question 2

Let $f$ be a function such that its derivative is $f'(x) = (x-a)^n g(x)$ , where $n$ is a positive integer, $a$ is a constant, and $g$ is a continuous function with $g(a) > 0$ . Under which condition will $f$ have a relative minimum at $x=a$ ?

If $n$ is any even positive integer
If $n$ is any odd positive integer (correct answer)
For all positive integers $n$
If $g'(a)$ exists and is negative

Explanation: When you encounter a problem about relative extrema using the first derivative, you need to analyze how the sign of

f'(x)

changes around the critical point where

f'(a) = 0

. Since

f'(x) = (x-a)^n g(x)

and

g(a) > 0

with

g

continuous, the function

g(x)

remains positive in some neighborhood around

x = a

. This means the sign of

f'(x)

near

x = a

depends entirely on the factor

(x-a)^n

. For a relative minimum, you need

f'(x)

to change from negative to positive as

x

increases through

a

. When

n

is odd,

(x-a)^n

is negative for

x < a

and positive for

x > a

. Since

g(x) > 0

throughout this interval,

f'(x)

changes from negative to positive, creating a relative minimum at

x = a

. Choice A is incorrect because when

n

is even,

(x-a)^n

is positive on both sides of

x = a

, so

f'(x)

doesn't change sign—there's no extremum. Choice C is wrong because, as explained above, even values of

n

don't produce relative minima. Choice D is irrelevant because the behavior depends on the power

n

, not on properties of

g'(a)

. The continuity of

g

and the fact that

g(a) > 0

are sufficient. Study tip: Remember that for critical points of the form

(x-a)^n

, odd powers create extrema while even powers create neither maxima nor minima (they're inflection points or flat regions).

Question 3

Let $f(x) = (x^2-4)^{2/3}$ . Which of the following statements correctly describes all the relative extrema of $f$ ?

$f$ has no relative extrema because its derivative is undefined at some points.
$f$ has a relative minimum at $x=0$ and relative maxima at $x=-2$ and $x=2$ .
$f$ has only a relative maximum at $x=0$ .
$f$ has a relative maximum at $x=0$ and relative minima at $x=-2$ and $x=2$ . (correct answer)

Explanation: When finding relative extrema, you need to locate critical points where the derivative equals zero or is undefined, then determine whether each point is a maximum, minimum, or neither using the first or second derivative test. To find the critical points of

f(x) = (x^2-4)^{2/3}

, first compute the derivative using the chain rule:

f'(x) = \frac{2}{3}(x^2-4)^{-1/3} \cdot 2x = \frac{4x}{3(x^2-4)^{1/3}}

The derivative equals zero when

x = 0

, and it's undefined when

x^2-4 = 0

, giving us

x = \pm 2

. So the critical points are

x = -2, 0, 2

. To classify these points, examine the sign of

f'(x)

around each critical point. For

x < -2

f'(x) < 0

; for

-2 < x < 0

f'(x) < 0

; for

0 < x < 2

f'(x) > 0

; for

x > 2

f'(x) > 0

. This means

f

decreases then increases at

x = 0

(relative minimum at each of

x = \pm 2

), and increases then decreases... wait, let me recalculate. Actually,

f

increases approaching

x = 0

from the left and decreases leaving

x = 0

to the right, making

x = 0

a relative maximum. At

x = \pm 2

, the function reaches its minimum value of 0. Answer D correctly identifies that

f

has a relative maximum at

x = 0

and relative minima at

x = \pm 2

. Answer A is wrong because undefined derivatives can still indicate extrema. Answer B incorrectly swaps the classifications. Answer C misses the minima at

x = \pm 2

. Study tip: Always check critical points where derivatives are undefined—they often yield important extrema, especially with fractional exponents.

Question 4

A function $f(x)$ is differentiable and has a critical point at $x=2$ . Which of the following conditions is NOT sufficient to determine whether $f$ has a relative maximum, a relative minimum, or neither at $x=2$ by using the logic of the First Derivative Test?

$f'(1) = 3$ and $f'(3) = -2$ .
$f'(x) = (x-2)g(x)$ , where the function $g(x)$ is positive for all $x$ .
$f(1) = 5$ , $f(2) = 3$ , and $f(3) = 5$ . (correct answer)
$f'(x)<0$ for all $x$ in $(0, 2)$ and $f'(x)>0$ for all $x$ in $(2, 4)$ .

Explanation: The First Derivative Test determines the type of extremum at a critical point

c

by examining the sign of the derivative,

f'(x)

, on either side of

c

. A condition is sufficient if it provides this information about the sign of

f'(x)

. A)

f'(1)=3

(positive) and

f'(3)=-2

(negative). This shows that

f'(x)

changes from positive to negative around

x=2

. This is sufficient to conclude

f

has a relative maximum at

x=2

. B)

f'(x) = (x-2)g(x)

with

g(x)>0

. For

x<2

(x-2)

is negative, so

f'(x)

is negative. For

x>2

(x-2)

is positive, so

f'(x)

is positive. This shows

f'(x)

changes from negative to positive. This is sufficient to conclude

f

has a relative minimum at

x=2

. C)

f(1)=5

f(2)=3

, and

f(3)=5

. This provides values of the function

f(x)

, not its derivative

f'(x)

. While these values suggest a relative minimum may exist at

x=2

, it is not guaranteed. For example,

f(x)

could decrease between

x=2

and

x=3

before increasing to

f(3)=5

. The First Derivative Test relies explicitly on the sign of the derivative, which is not given here. Therefore, this information is not sufficient. D) This states that

f'(x)

is negative to the left of 2 and positive to the right of 2. This is the definition of the condition for a relative minimum according to the First Derivative Test. This is sufficient.

Question 5

A function f is twice-differentiable. It is known that f'(c)=0. To determine if f has a local extremum at x=c using the First Derivative Test, which of the following is sufficient?

f''(c) > 0.
f'(x) changes sign at x=c. (correct answer)
f'(x) > 0 for x > c and f'(x) < 0 for x < c.
f(x) is continuous at x=c.

Explanation: The First Derivative Test for a local extremum at a critical point c states that if the derivative f'(x) changes sign at x=c, then f has a local extremum at c. Choice A is the condition for the Second Derivative Test, not the First. Choice C describes the conditions for a local minimum specifically, which is sufficient for an extremum, but it is not the most general condition for an extremum (which could be a maximum or a minimum). Choice B is the core condition of the test for the existence of any local extremum. Choice D is a necessary condition for differentiability at c (which implies f'(c) exists), but it is not sufficient to guarantee an extremum.

Question 6

Let $f(x) = x(x-4)^3$ . If $g(x) = |f(x)|$ , at which x-values does $g(x)$ have relative minima?

At $x=1$ and $x=4$ only
At $x=1$ only
At $x=0$ and $x=4$ only (correct answer)
At $x=0, x=1$ , and $x=4$

Explanation: When you see absolute value functions and need to find relative minima, you're looking for points where the function changes from decreasing to increasing. The key insight is that

g(x) = |f(x)|

will have relative minima wherever

f(x) = 0

(creating sharp corners) or wherever

f(x)

itself has relative minima that remain below the x-axis. First, let's analyze

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

. To find critical points, we need

f'(x) = (x-4)^3 + x \cdot 3(x-4)^2 = (x-4)^2[(x-4) + 3x] = (x-4)^2(4x-4) = 4(x-4)^2(x-1)

. Setting

f'(x) = 0

gives us

x = 1

and

x = 4

. Next, find where

f(x) = 0

x(x-4)^3 = 0

gives us

x = 0

and

x = 4

. These are crucial because

|f(x)|

creates sharp points here. Now analyze the behavior:

f(x) < 0

when

0 < x < 4

, and

f(x) > 0

when

x < 0

x > 4

. At

x = 0

and

x = 4

, the function

g(x) = |f(x)|

touches zero and changes direction, creating relative minima. At

x = 1

, even though

f(x)

has a relative maximum, since

f(1) = 1(-3)^3 = -27 < 0

, we have

g(1) = 27

, which is actually a relative maximum for

g(x)

, not a minimum. Choice A incorrectly includes

x = 1

. Choice B only identifies

x = 1

, missing the actual minima. Choice D includes all critical points without considering the behavior of the absolute value. Strategy tip: For absolute value problems, always identify where the inside function equals zero and check whether critical points create maxima or minima after applying absolute value.

Question 7

Let $g(x) = (x^2 - 4)^{2/3}$ . Which of the following correctly describes all relative extrema of the function $g$ ?

Relative minima at $x=-2$ and $x=2$ and no other relative extrema.
A relative minimum at $x=0$ and relative maxima at $x=-2$ and $x=2$ .
A relative maximum at $x=0$ and no other relative extrema.
A relative maximum at $x=0$ and relative minima at $x=-2$ and $x=2$ . (correct answer)

Explanation: To find relative extrema, you need to locate critical points where the derivative equals zero or is undefined, then determine whether each point is a maximum, minimum, or neither. First, find the derivative using the chain rule:

g'(x) = \frac{2}{3}(x^2 - 4)^{-1/3} \cdot 2x = \frac{4x}{3(x^2 - 4)^{1/3}}

Critical points occur when

g'(x) = 0

or is undefined. The derivative equals zero when

x = 0

, and it's undefined when

x^2 - 4 = 0

, giving us

x = -2

and

x = 2

. Now analyze the behavior around each critical point. For

x = 0

: when

x < 0

, we have

g'(x) < 0

(negative slope), and when

x > 0

, we have

g'(x) > 0

(positive slope). This indicates a relative minimum at

x = 0

. Wait - let me recalculate this more carefully. At

x = 0

g'(x) = \frac{4x}{3(-4)^{1/3}} = \frac{4x}{3(-4)^{1/3}}

. For

x

slightly less than 0,

g'(x) > 0

, and for

x

slightly greater than 0,

g'(x) < 0

. This means we have a relative maximum at

x = 0

. At

x = -2

and

x = 2

, the derivative is undefined, but the function value is

g(-2) = g(2) = 0

, which are the lowest points on the graph, making them relative minima. Choice A ignores the critical point at

x = 0

. Choice B incorrectly identifies the nature of all three extrema. Choice C misses the minima at

x = ±2

. Study tip: When derivatives are undefined due to fractional exponents, don't forget to check if those points are still extrema by examining the function's behavior around them.

Question 8

Let $f(x) = \frac{\ln x}{x^2}$ for $x>0$ . At which value of $x$ does $f(x)$ have a relative extremum?

A relative maximum at $x = \sqrt{e}$ (correct answer)
A relative minimum at $x = \sqrt{e}$
A relative maximum at $x = e^2$
A relative minimum at $x = 1$

Explanation: First, find the derivative using the quotient rule:

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

. To find critical points, set

f'(x) = 0

. Since

x>0

, we only need to solve

1 - 2 \ln x = 0

, which gives

2 \ln x = 1

, or

\ln x = 1/2

, so

x = e^{1/2} = \sqrt{e}

. To classify this extremum, we use the First Derivative Test. For

0 < x < \sqrt{e}

, choose

x=1

f'(1) = \frac{1-2\ln 1}{1^3} = 1 > 0

. For

x > \sqrt{e}

, choose

x=e

f'(e) = \frac{1-2\ln e}{e^3} = \frac{1-2}{e^3} < 0

. Since the derivative changes from positive to negative,

f

has a relative maximum at

x = \sqrt{e}

Question 9

The derivative of a function $f$ is given by $f'(x) = (x-k)(x+2)$ . The function $f$ is known to have a relative minimum at $x=3$ . What is the nature of the function's other critical point at $x=-2$ ?

A relative maximum (correct answer)
A relative minimum
A critical point, but not a relative extremum
The point $x=-2$ is not a critical point

Explanation: For

f

to have a relative minimum at

x=3

, its derivative

f'(x)

must be zero at

x=3

and change sign from negative to positive. The critical points of

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

are

x=k

and

x=-2

. Since one of the critical points is at

x=3

, we must have

k=3

. Thus,

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

. The critical points are

x=3

and

x=-2

. Let's test the sign of

f'(x)

around

x=-2

. For

x<-2

(e.g.,

x=-3

f'(-3) = (-6)(-1) > 0

. For

-2<x<3

(e.g.,

x=0

f'(0) = (-3)(2) < 0

. Since

f'(x)

changes from positive to negative at

x=-2

, the function

f

has a relative maximum at

x=-2

Question 10

Let $f(x) = x^4 - 4x^3 + 4x^2 + 1$ . Which of the following statements correctly classifies all the relative extrema of $f$ ?

Relative maxima at $x=0$ and $x=2$ ; a relative minimum at $x=1$ .
Relative minima at $x=0$ and $x=2$ ; a relative maximum at $x=1$ . (correct answer)
A relative maximum at $x=1$ ; no relative minima.
A relative minimum at $x=1$ ; no relative maxima.

Explanation: When you encounter a question about classifying relative extrema, you need to find critical points using the first derivative, then determine whether each is a maximum or minimum using the second derivative test. Start by finding

f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x-1)(x-2)

. Setting this equal to zero gives critical points at

x = 0, 1, 2

. Next, use the second derivative:

f''(x) = 12x^2 - 24x + 8

. Now test each critical point:

At $x = 0$ : $f''(0) = 8 > 0$ , so this is a relative minimum
At $x = 1$ : $f''(1) = 12 - 24 + 8 = -4 < 0$ , so this is a relative maximum
At $x = 2$ : $f''(2) = 48 - 48 + 8 = 8 > 0$ , so this is a relative minimum

This confirms answer B: relative minima at

x = 0

and

x = 2

, with a relative maximum at

x = 1

. Answer A incorrectly swaps the classifications—it claims maxima where there are actually minima and vice versa. This often happens when students confuse the second derivative test rule. Answer C ignores the minima at

x = 0

and

x = 2

, while answer D completely reverses the situation by claiming no maxima exist when there's clearly one at

x = 1

. Remember: when

f''(x) > 0

at a critical point, you have a relative minimum (the graph is concave up, like a bowl). When

f''(x) < 0

, you have a relative maximum (concave down, like an upside-down bowl).

Question 11

The derivative of a function $g$ is given by $g'(x) = (x^3-4x)\ln(x^2+1)$ . At which of the following x-values does $g$ have a relative maximum?

$x=-2$ and $x=2$
$x=2$ only
$x=0$ only (correct answer)
$x=-2, 0,$ and $2$

Explanation: When you're asked to find relative maxima given a derivative, you need to locate critical points where the derivative equals zero, then use the second derivative test or analyze sign changes to determine whether each critical point is a maximum, minimum, or neither. First, find where

g'(x) = (x^3-4x)\ln(x^2+1) = 0

. Since

\ln(x^2+1)

is always positive (because

x^2+1 \geq 1

for all real

x

), we only need

(x^3-4x) = 0

. Factoring gives

x(x^2-4) = x(x-2)(x+2) = 0

, so the critical points are

x = -2, 0, 2

. To determine which are maxima, analyze how

g'(x)

changes sign around each critical point. The factor

\ln(x^2+1)

is always positive, so the sign depends entirely on

(x^3-4x)

For $x < -2$ : $(x^3-4x) < 0$ , so $g'(x) < 0$
For $-2 < x < 0$ : $(x^3-4x) > 0$ , so $g'(x) > 0$
For $0 < x < 2$ : $(x^3-4x) < 0$ , so $g'(x) < 0$
For $x > 2$ : $(x^3-4x) > 0$ , so $g'(x) > 0$

A relative maximum occurs when the derivative changes from positive to negative. This happens only at

x = 0

(positive to negative). At

x = -2

and

x = 2

, the derivative changes from negative to positive, indicating relative minima. Choice A incorrectly identifies the minima as maxima. Choice B misses that

x = 2

is a minimum, not a maximum. Choice D incorrectly includes all critical points without checking the sign changes. Remember: always verify whether critical points are maxima or minima by checking how the derivative's sign changes around each point.

Question 12

Let $f(x) = x^2 e^{-ax}$ where $a$ is a positive constant. The function $f$ has a relative extremum for some $x>0$ . What is the location and type of this extremum?

A relative maximum at $x = 2/a$ . (correct answer)
A relative minimum at $x = 2/a$ .
A relative minimum at $x = 0$ .
A relative maximum at $x = a/2$ .

Explanation: To find the relative extrema, we first find the derivative of

f(x)

using the product rule:

f'(x) = (2x)(e^{-ax}) + (x^2)(-ae^{-ax})

. Factor out common terms:

f'(x) = xe^{-ax}(2-ax)

. Next, find the critical points by setting

f'(x)=0

. Since

e^{-ax}

is always positive, we only need to solve

x(2-ax)=0

. This gives

x=0

and

2-ax=0

, which means

x=2/a

. The problem specifies

x>0

, so we are interested in the critical point

x=2/a

. (Since

a>0

2/a

is positive). Now, we apply the First Derivative Test to the point

x=2/a

. We analyze the sign of

f'(x) = xe^{-ax}(2-ax)

in an interval around

2/a

. For

x>0

, both

x

and

e^{-ax}

are positive, so the sign of

f'(x)

is determined by the sign of the term

(2-ax)

When $0 < x < 2/a$ , we have $ax < 2$ , so $2-ax > 0$ . Thus, $f'(x)$ is positive.
When $x > 2/a$ , we have $ax > 2$ , so $2-ax < 0$ . Thus, $f'(x)$ is negative.

Since

f'(x)

changes from positive to negative at

x=2/a

, the function

f(x)

has a relative maximum at this point.

Question 13

The derivative of a function $f$ is given by $f'(x) = (x-a)^2 (x-b)$ for constants $a$ and $b$ where $a \neq b$ . Which of the following statements must be true about the relative extrema of $f$ ?

$f$ has a relative extremum at $x=a$ .
$f$ has a relative maximum at $x=b$ .
$f$ has a relative minimum at $x=b$ . (correct answer)
The type of extremum at $x=b$ depends on whether $a>b$ or $b>a$ .

Explanation: When you encounter a derivative and need to analyze extrema, you're looking for critical points where the derivative equals zero, then determining whether each critical point is a maximum, minimum, or neither using the first or second derivative test. Given

f'(x) = (x-a)^2(x-b)

, the critical points occur when

f'(x) = 0

. This happens when

(x-a)^2 = 0

(x-b) = 0

, giving us critical points at

x = a

and

x = b

. To determine the nature of each critical point, let's analyze the sign of

f'(x)

around these points. Since

(x-a)^2

is always non-negative (it's a perfect square), the sign of

f'(x)

depends entirely on the sign of

(x-b)

. When

x < b

, we have

(x-b) < 0

, so

f'(x) \leq 0

. When

x > b

, we have

(x-b) > 0

, so

f'(x) \geq 0

. This means

f'(x)

changes from negative to positive at

x = b

, indicating a relative minimum there. Now examining each choice: (A) is incorrect because at

x = a

, the factor

(x-a)^2

equals zero, but since it's squared,

f'(x)

doesn't change sign around

x = a

– it remains the same sign as

(x-b)

on both sides. (B) is wrong because we showed

x = b

gives a minimum, not maximum. (D) is incorrect because the relative minimum at

x = b

occurs regardless of whether

a > b

b > a

. Study tip: When analyzing critical points, always check whether the derivative actually changes sign – squared factors don't cause sign changes, so they don't create extrema.

Question 14

Let $g(x) = \int_{0}^{x} (t^2-4t+3)e^{-t^2} dt$ . At which of the following values of $x$ does $g(x)$ have a relative maximum?

$x=0$
$x=1$ (correct answer)
$x=2$
$x=3$

Explanation: According to the Fundamental Theorem of Calculus, Part 1, the derivative of

g(x)

is given by

g'(x) = (x^2-4x+3)e^{-x^2}

. To find the critical points of

g(x)

, we set

g'(x)=0

(x^2-4x+3)e^{-x^2} = 0

. Since

e^{-x^2}

is always positive, we only need to solve

x^2-4x+3 = 0

. Factoring the quadratic gives

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

. The critical points are

x=1

and

x=3

. Now, we use the First Derivative Test to classify these critical points. The sign of

g'(x)

is determined by the sign of the quadratic factor

(x-1)(x-3)

, which represents an upward-opening parabola.

For $x<1$ , $(x-1)$ is negative and $(x-3)$ is negative, so their product is positive. Thus, $g'(x)>0$ .
For $1<x<3$ , $(x-1)$ is positive and $(x-3)$ is negative, so their product is negative. Thus, $g'(x)<0$ .
For $x>3$ , $(x-1)$ is positive and $(x-3)$ is positive, so their product is positive. Thus, $g'(x)>0$ .

x=1

, the derivative

g'(x)

changes from positive to negative, which indicates a relative maximum. At

x=3

, the derivative

g'(x)

changes from negative to positive, which indicates a relative minimum. The question asks for the location of the relative maximum, which is at

x=1

Question 15

Let $f$ be a function with derivative $f'(x) = (x^2 - kx)(x-4)$ , where $k$ is a constant. If $f$ has a relative maximum at $x=2$ , which of the following statements about $f$ is true?

$f$ has a relative minimum at $x=0$ . (correct answer)
$f$ has a relative maximum at $x=4$ .
$f$ has a relative minimum at $x=4$ .
$f$ has a relative maximum at $x=0$ .

Explanation: For

f

to have a relative extremum at

x=2

x=2

must be a critical point, so

f'(2)=0

. Substitute

x=2

into the expression for

f'(x)

f'(2) = (2^2 - k(2))(2-4) = (4-2k)(-2) = 0

. Solving for

k

gives

4-2k=0

, so

k=2

. Now, substitute

k=2

back into the derivative:

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

. The critical points are the roots of

f'(x)

, which are

x=0

x=2

, and

x=4

. To determine the nature of the extrema, we use the First Derivative Test. We test the sign of

f'(x)

in the intervals defined by the critical points:

For $x<0$ (e.g., $x=-1$ ), $f'(-1) = (-1)(-3)(-5) < 0$ . $f$ is decreasing.
For $0<x<2$ (e.g., $x=1$ ), $f'(1) = (1)(-1)(-3) > 0$ . $f$ is increasing.
For $2<x<4$ (e.g., $x=3$ ), $f'(3) = (3)(1)(-1) < 0$ . $f$ is decreasing.
For $x>4$ (e.g., $x=5$ ), $f'(5) = (5)(3)(1) > 0$ . $f$ is increasing.

x=0

, the derivative changes from negative to positive, so

f

has a relative minimum at

x=0

. At

x=2

, the derivative changes from positive to negative, confirming the given relative maximum. At

x=4

, the derivative changes from negative to positive, so

f

has a relative minimum at

x=4

. Therefore, the statement that

f

has a relative minimum at

x=0

is true.

Question 16

Consider the function $f(x) = \frac{x^2+3}{x-1}$ . Which of the following statements correctly describes the relative extrema of $f$ ?

$f$ has a relative minimum at $x=1$ and a relative maximum at $x=3$ .
$f$ has a relative minimum at $x=-1$ and a relative maximum at $x=3$ .
$f$ has a relative maximum at $x=-1$ , a relative minimum at $x=3$ , and a relative extremum at $x=1$ .
$f$ has a relative maximum at $x=-1$ and a relative minimum at $x=3$ . (correct answer)

Explanation: When finding relative extrema for rational functions, you need to locate critical points by finding where the derivative equals zero or is undefined, then analyze the behavior around those points. For

f(x) = \frac{x^2+3}{x-1}

, let's find

f'(x)

using the quotient rule:

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

Setting the numerator equal to zero:

x^2-2x-3 = 0

, which factors as

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

. This gives critical points at

x = -1

and

x = 3

. Note that

x = 1

makes the derivative undefined, but it's also where the original function is undefined (vertical asymptote), so it's not a candidate for a relative extremum. Testing the sign of

f'(x)

around our critical points: for

x < -1

f'(x) > 0

; for

-1 < x < 1

f'(x) < 0

; for

1 < x < 3

f'(x) < 0

; for

x > 3

f'(x) > 0

. Since

f'(x)

changes from positive to negative at

x = -1

, this is a relative maximum. Since

f'(x)

changes from negative to positive at

x = 3

, this is a relative minimum. Choice A is wrong because

x = 1

is not in the domain. Choice B incorrectly identifies

x = -1

as a minimum. Choice C incorrectly claims there's an extremum at the undefined point

x = 1

. Choice D correctly identifies the relative maximum at

x = -1

and relative minimum at

x = 3

. Remember: always check that critical points are actually in the function's domain before classifying them as extrema.

Question 17

Let $f(x) = x - 2\cos(x)$ on the interval $[0, 2\pi]$ . At which value of $x$ does $f(x)$ have a relative minimum?

$x=\frac{7\pi}{6}$
$x=\frac{11\pi}{6}$ (correct answer)
$x=\frac{\pi}{6}$
$x=\frac{5\pi}{3}$

Explanation: To find relative extrema, we first compute the derivative of

f(x)

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

. Next, we find the critical points by setting

f'(x)=0

for

x

[0, 2\pi]

1 + 2\sin(x) = 0

2\sin(x) = -1

\sin(x) = -\frac{1}{2}

The solutions for

x

in the interval

[0, 2\pi]

are

x = \frac{7\pi}{6}

and

x = \frac{11\pi}{6}

. Now we use the First Derivative Test to classify these critical points. We check the sign of

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

on the intervals created by these points:

[0, \frac{7\pi}{6})

(\frac{7\pi}{6}, \frac{11\pi}{6})

, and

(\frac{11\pi}{6}, 2\pi]

Interval $[0, \frac{7\pi}{6})$ : Let's test $x=\pi$ . $f'(\pi) = 1 + 2\sin(\pi) = 1+0=1 > 0$ .
Interval $(\frac{7\pi}{6}, \frac{11\pi}{6})$ : Let's test $x=\frac{3\pi}{2}$ . $f'(\frac{3\pi}{2}) = 1 + 2\sin(\frac{3\pi}{2}) = 1+2(-1) = -1 < 0$ .
Interval $(\frac{11\pi}{6}, 2\pi]$ : Let's test $x=2\pi$ . $f'(2\pi) = 1 + 2\sin(2\pi) = 1+0=1 > 0$ .

x=\frac{7\pi}{6}

f'(x)

changes from positive to negative, indicating a relative maximum. At

x=\frac{11\pi}{6}

f'(x)

changes from negative to positive, indicating a relative minimum. The question asks for the relative minimum, which occurs at

x=\frac{11\pi}{6}

Question 18

Let f(x) = |x^2 - 4x|. Which of the following statements correctly describes the local extrema of f?

f has local minima at x=0 and x=4, and a local maximum at x=2. (correct answer)
f has local maxima at x=0 and x=4, and a local minimum at x=2.
f has local minima at x=0, x=2, and x=4.
f has a local minimum at x=2 and local maxima at x=0 and x=4.

Explanation: We can write f(x) as a piecewise function. Since x^2-4x = x(x-4), the expression is non-negative for x \leq 0 or x \geq 4, and negative for 0 < x < 4. So, f(x) = x^2-4x for x \in (-

\infty

, 0] \cup [4,

\infty

), and f(x) = -(

x^2-4x

) = 4x-x^2 for x \in (0, 4). The derivatives are f'(x) = 2x-4 for x<0 or x>4, and f'(x) = 4-2x for 0<x<4. Critical points occur where f'(x)=0 or f'(x) is undefined. f'(x) is undefined at x=0 and x=4 (sharp corners). Setting 4-2x=0 gives x=2, which is in (0,4). So critical points are x=0, 2, 4. For x<0, f' is negative. For 0<x<2, f' = 4-2x is positive. At x=0, f' changes from negative to positive, so it's a local minimum. For 2<x<4, f' = 4-2x is negative. At x=2, f' changes from positive to negative, so it's a local maximum. For x>4, f' = 2x-4 is positive. At x=4, f' changes from negative to positive, so it's a local minimum.

Question 19

The function f(x) = x^{2/3}(x-5) has two critical points. What is the nature of these points?

One local maximum and one local minimum. (correct answer)
Two local minima.
Two local maxima.
One local minimum and one point that is not an extremum.

Explanation: First, find the derivative. f(x) = x^{5/3} - 5x^{2/3}. So, f'(x) = (5/3)x^{2/3} - (10/3)x^{-1/3}. We can factor this as f'(x) = (5/3)x^{-1/3}(x - 2). Critical points occur when f'(x) = 0 or f'(x) is undefined. f'(x) = 0 when x=2. f'(x) is undefined when x=0. The critical points are x=0 and x=2. We test the sign of f'(x) = \frac{5(x-2)}{3x^{1/3}}. For x<0, numerator is negative, denominator is negative, so f'(x) is positive. For 0<x<2, numerator is negative, denominator is positive, so f'(x) is negative. For x>2, numerator is positive, denominator is positive, so f'(x) is positive. At x=0, f' changes from positive to negative, indicating a local maximum. At x=2, f' changes from negative to positive, indicating a local minimum.

Question 20

The function f(x) = ax^4 - 4x^3 + 1 has a local minimum at x=1. What is the nature of the function's other critical point?

A local maximum.
A local minimum.
An inflection point that is not an extremum. (correct answer)
The function has no other critical point.

Explanation: First, find the derivative: f'(x) = 4ax^3 - 12x^2. Since f has a local minimum at x=1, it must be a critical point, so f'(1) = 0. 4a(1)^3 - 12(1)^2 = 0 \implies 4a - 12 = 0 \implies a = 3. Now substitute a=3 back into the derivative: f'(x) = 12x^3 - 12x^2 = 12x^2(x-1). The critical points are where f'(x)=0, so x=0 and x=1. The problem gives us x=1. We must analyze the other critical point, x=0. To do this, we check the sign of f'(x) around x=0. The term 12x^2 is always non-negative. For x < 0, (x-1) is negative, so f'(x) is negative. For 0 < x < 1, (x-1) is negative, so f'(x) is also negative. Since the sign of f'(x) does not change at x=0, it is not a local extremum. It is a critical point that is also an inflection point.