Calculus 1 Quiz: Evt Extrema And Critical Points

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Evt Extrema And Critical PointsQuestion 1 of 20

What is the absolute minimum value of the function $f(x) = 2x^3 - 9x^2 + 12x + 1$ on the interval $[0, 3]$ ?

Calculus 1 Quiz

Calculus 1 Quiz: Evt Extrema And Critical Points

Practice Evt Extrema And Critical Points in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Evt Extrema And Critical Points, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

What is the absolute minimum value of the function $f(x) = 2x^3 - 9x^2 + 12x + 1$ on the interval $[0, 3]$ ?

1 (correct answer)
5
6
10

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

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)

. The critical points are

x=1

and

x=2

, both of which are in the interval

[0, 3]

. To find the absolute minimum, we evaluate the function at the critical points and the endpoints:

f(0) = 1

f(1) = 6

f(2) = 5

, and

f(3) = 10

. Comparing these values, the smallest is 1. Thus, the absolute minimum value is 1.

Question 2

How many critical points does the function $f(x) = (x^2 - 4)^{\frac{2}{3}}$ have?

1
2
3 (correct answer)
0

Explanation: To find critical points, we compute the derivative:

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

. Critical points occur where

f'(x) = 0

f'(x)

is undefined.

f'(x) = 0

when the numerator is zero, so

4x=0 \implies x=0

f'(x)

is undefined when the denominator is zero, so

x^2-4=0 \implies x=2

and

x=-2

. Since the original function

f(x)

is defined at all three of these points, there are a total of 3 critical points.

Question 3

Let $f(x)$ be defined as:

Which statement accurately describes the extrema of $f$ on the interval $[-2, 4]$ ?

The absolute maximum is 4 and the absolute minimum is 0.
The absolute maximum is 6 and the absolute minimum is 0. (correct answer)
The absolute maximum is 4 and the absolute minimum is 3.
The absolute maximum is 6 and the absolute minimum is 3.

Explanation: We must check the critical points and the endpoints of the interval

[-2, 4]

. For

x < 1

f'(x) = -2x

, so

x=0

is a critical point. For

x > 1

f'(x) = 1

, so there are no critical points. At

x=1

, the function is continuous, but the left-hand derivative is -2 and the right-hand derivative is 1, so

x=1

is a critical point (a corner). The candidates for absolute extrema are the endpoints

-2, 4

and the critical points

0, 1

. Evaluating the function at these points:

f(-2)=0

f(0)=4

f(1)=3

f(4)=6

. Comparing these values, the absolute maximum is 6 and the absolute minimum is 0.

Question 4

A particle moves along the x-axis with position given by $s(t) = \frac{1}{4}t^4 - t^3 - 2t^2 + 1$ for the time interval $0 \leq t \leq 5$ . At what time $t$ is the particle furthest from the origin in the negative direction?

$t=0$
$t=4$ (correct answer)
$t=5$
$t=\frac{3+\sqrt{17}}{2}$

Explanation: The phrase "furthest from the origin in the negative direction" means we are looking for the absolute minimum value of the position function

s(t)

on the interval

[0, 5]

. First, find the derivative of

s(t)

, which represents the velocity

v(t)

v(t) = s'(t) = t^3 - 3t^2 - 4t

. Next, find the critical points by setting the derivative to zero:

t^3 - 3t^2 - 4t = 0

t(t^2 - 3t - 4) = 0

t(t-4)(t+1) = 0

The critical times are

t=0, t=4

, and

t=-1

. We only consider the critical points within the given interval

[0, 5]

, which are

t=0

and

t=4

. To find the absolute minimum, we test these points and the endpoint

t=5

in the position function

s(t)

s(0) = \frac{1}{4}(0)^4 - (0)^3 - 2(0)^2 + 1 = 1

s(4) = \frac{1}{4}(4)^4 - (4)^3 - 2(4)^2 + 1 = 64 - 64 - 32 + 1 = -31

s(5) = \frac{1}{4}(5)^4 - (5)^3 - 2(5)^2 + 1 = \frac{625}{4} - 125 - 50 + 1 = 156.25 - 174 = -17.75

Comparing the values

s(0)=1

s(4)=-31

, and

s(5)=-17.75

, the most negative position (the absolute minimum) is

-31

, which occurs at

t=4

Question 5

Let $f$ be a continuous function on the closed interval $[0, 5]$ . The only critical point of $f$ in the open interval $(0, 5)$ is at $x=2$ . If $f(0)=1, f(2)=4,$ and $f(5)=3$ , what can be concluded about the absolute extrema of $f$ on $[0, 5]$ ?

The absolute minimum is 1 and the absolute maximum is 4. (correct answer)
The absolute minimum is 1 and the absolute maximum is 3.
The absolute maximum is 4, but the absolute minimum cannot be determined.
Neither the absolute maximum nor the absolute minimum can be determined from the information given.

Explanation: According to the Extreme Value Theorem, a continuous function on a closed interval must have an absolute maximum and an absolute minimum. These extrema can only occur at critical points within the interval or at the endpoints of the interval. The possible locations for the absolute extrema are the endpoints

x=0

and

x=5

, and the critical point

x=2

. We are given the function values at these three candidate points:

f(0) = 1

f(2) = 4

f(5) = 3

To find the absolute maximum and minimum on the interval

[0, 5]

, we simply compare these values. The largest value is 4, which occurs at

x=2

. So, the absolute maximum is 4. The smallest value is 1, which occurs at

x=0

. So, the absolute minimum is 1. Because the problem states that

x=2

is the only critical point in the open interval, we have a complete list of all possible locations for the absolute extrema. Therefore, we can definitively determine both.

Question 6

Let $f$ be a function that is defined for all real numbers. If $f'(2)$ is undefined, which of the following statements must be true?

$f$ must have a local extremum at $x=2$ .
$f$ must have a critical point at $x=2$ . (correct answer)
$f$ must be discontinuous at $x=2$ .
The graph of $f$ must have a vertical tangent at $x=2$ .

Explanation: The definition of a critical point of a function

f

is a point

x=c

in the domain of

f

where either

f'(c)=0

f'(c)

is undefined. The stem states that

f

is defined for all real numbers (so

x=2

is in the domain) and that

f'(2)

is undefined. Therefore, by definition,

f

must have a critical point at

x=2

. (A) is false. A critical point is a necessary condition for a local extremum, but not a sufficient one. For example, if

f(x) = (x-2)^{1/3}

, then

f'(2)

is undefined, but

f

has no extremum at

x=2

. (C) is false. A function can be continuous at a point where its derivative is undefined. For example,

f(x) = |x-2|

is continuous at

x=2

, but

f'(2)

is undefined. (D) is false. An undefined derivative can correspond to a corner or a cusp, not just a vertical tangent. For example,

f(x)=|x-2|

has a corner at

x=2

Question 7

Let $f(x)$ be a piecewise function defined as $f(x) = \begin{cases} x^2+4x+3 & \text{if } x \leq -1 \\ -x-1 & \text{if } x > -1 \end{cases}$ . What is the sum of the x-coordinates of all local extrema of $f$ ?

$-1$
$-2$
$-3$ (correct answer)
The function has no local extrema.

Explanation: To find local extrema, we must find all critical points of

f

. We check where

f'(x)=0

and where

f'(x)

is undefined.

Differentiate each piece: For $x < -1$ , $f'(x) = 2x+4$ . Setting this to zero gives $2x+4=0 \implies x=-2$ . Since $-2 < -1$ , this is a valid critical point. For $x > -1$ , $f'(x) = -1$ . This is never zero, so there are no critical points in this region.
Check the point where the definition changes, $x=-1$ . A critical point exists here if the derivative is undefined. First, check for continuity: $\lim_{x \to -1^-} (x^2+4x+3) = (-1)^2+4(-1)+3 = 0$ . $\lim_{x \to -1^+} (-x-1) = -(-1)-1 = 0$ . Since the limits are equal, the function is continuous. Now check the one-sided derivatives: Left-hand derivative: $f'(-1^-) = 2(-1)+4 = 2$ . Right-hand derivative: $f'(-1^+) = -1$ . Since the left-hand and right-hand derivatives are not equal, $f'(-1)$ is undefined. Thus, $x=-1$ is a critical point.
The critical points are $x=-2$ and $x=-1$ . We classify them using the first derivative test:
- For $x<-2$ , $f'(x)$ is negative (e.g., $f'(-3)=-2$ ). For $-2<x<-1$ , $f'(x)$ is positive (e.g., $f'(-1.5)=1$ ). Thus, $f$ has a local minimum at $x=-2$ .
- For $-2<x<-1$ , $f'(x)$ is positive. For $x>-1$ , $f'(x)$ is negative (it's -1). Thus, $f$ has a local maximum at $x=-1$ .

Both critical points correspond to local extrema. The sum of their x-coordinates is

(-2) + (-1) = -3

Question 8

Let $f(x) = x\sqrt{4-x^2}$ on its domain $[-2, 2]$ . What is the difference between the absolute maximum and absolute minimum values of $f$ ?

0
2
4 (correct answer)
$2\sqrt{2}$

Explanation: First, find the derivative:

f'(x) = \sqrt{4-x^2} - \frac{x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}}

. Critical points occur when

4-2x^2=0

, which gives

x = \pm\sqrt{2}

. We must test these points and the endpoints

x=\pm 2

f(-2)=0

f(-\sqrt{2}) = -2

f(\sqrt{2}) = 2

f(2)=0

. The absolute maximum value is 2, and the absolute minimum value is -2. The difference is

2 - (-2) = 4

Question 9

A function $h(t)$ is continuous on $[0, 10]$ and represents the height of a rocket in kilometers $t$ minutes after launch. It is known that $h(0)=0$ , $h(10)=5$ , and the rocket reaches a maximum height of 8 km at some time during the flight. Let $t_{max}$ be the time of maximum height and $t_{min}$ be the time of minimum height. Which of the following must be true?

$t_{min}=0$ and $t_{max}$ must be in $(0, 10)$ . (correct answer)
$h'(t_{max})$ must be equal to 0.
$t_{min}$ must be 0, but $t_{max}$ could be an endpoint, 0 or 10.
Both the absolute minimum and maximum values must occur at critical points in $(0,10)$ .

Explanation: By the Extreme Value Theorem, absolute minimum and maximum values exist on the closed interval [0,10]. Since height cannot be negative and

h(0)=0

, the absolute minimum is 0, occurring at

t_{min}=0

. The maximum height is 8 km. Since

h(0)=0

and

h(10)=5

, both endpoint values are less than 8, so the maximum cannot occur at an endpoint. Therefore, the maximum must occur at some interior point, so

t_{max} \in (0, 10)

. Choice B is incorrect because we're only told

h

is continuous, not differentiable, so

h'(t_{max})

may not exist.

Question 10

Consider the function $g(x) = \tan(x)$ on the interval $[0, \pi]$ . Which of the following explains why the Extreme Value Theorem does not guarantee that $g(x)$ has an absolute maximum and an absolute minimum on this interval?

The interval $[0, \pi]$ is not a closed interval.
The function $g(x)$ is not differentiable on the interval $[0, \pi]$ .
The function $g(x)$ is not continuous on the interval $[0, \pi]$ . (correct answer)
The function $g(x)$ is not monotonic on the interval $[0, \pi]$ .

Explanation: The Extreme Value Theorem has two conditions: the function must be continuous and the interval must be closed. The interval

[0, \pi]

is closed. However,

g(x) = \tan(x)

has an infinite discontinuity at

x=\pi/2

, which lies within the interval. Because the function is not continuous on the entire interval, the theorem's conditions are not met, and it does not apply.

Question 11

Let $f$ be a function such that $f'(c) = 0$ . Which of the following additional conditions is sufficient to conclude that $f$ has a local minimum at $x=c$ ?

$f''(c)$ exists and is non-negative.
$f''(x)$ changes from negative to positive at $x=c$ .
$f'(x)$ changes from negative to positive at $x=c$ . (correct answer)
$f(c)$ is the smallest value of the function on some open interval containing $c$ .

Explanation: This question asks for a sufficient condition, which is a key part of a calculus test for extrema. The First Derivative Test states that if

f'(c)=0

and

f'(x)

changes sign from negative on the left of

c

to positive on the right of

c

, then

f

has a local minimum at

c

. This describes the function changing from decreasing to increasing. Choice D is the definition of a local minimum, not a test condition. Choice A is insufficient because if

f''(c)=0

, the test is inconclusive. Choice B describes a condition for an inflection point, not an extremum.

Question 12

A twice-differentiable function $f(x)$ is defined on $(-\infty, \infty)$ . It is known that $f(x)$ has exactly two critical points, at $x=-2$ and $x=5$ . If $f''(-2) > 0$ and $f''(5) < 0$ , which of the following must be true?

$f$ has a global minimum at $x=-2$ .
$f$ has a global maximum at $x=5$ .
$f(-2)$ is a local minimum value and $f(5)$ is a local maximum value. (correct answer)
$\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} f(x) = -\infty$ .

Explanation: The Second Derivative Test allows us to classify critical points. Since

f'(-2)=0

and

f''(-2)>0

f

has a local minimum at

x=-2

. Since

f'(5)=0

and

f''(5)<0

f

has a local maximum at

x=5

. This makes statement C correct. We cannot conclude anything about global extrema (A and B) because we do not know the end behavior of the function. For example, a function like

f(x) = -x^4 + ...

could satisfy these conditions but would have no global minimum. The end behavior described in D is not guaranteed either.

Question 13

What is the absolute maximum value of the function $g(x) = \sin(x) + \cos^2(x)$ on the interval $[0, \pi]$ ?

1
5/4 (correct answer)
3/2
2

Explanation: First, find the derivative:

g'(x) = \cos(x) - 2\cos(x)\sin(x) = \cos(x)(1 - 2\sin(x))

. Set

g'(x)=0

to find critical points in

[0, \pi]

. This occurs if

\cos(x)=0 \implies x=\pi/2

or if

1-2\sin(x)=0 \implies \sin(x)=1/2 \implies x=\pi/6, 5\pi/6

. Now evaluate

g(x)

at the critical points and endpoints:

g(0)=1

g(\pi/6) = 1/2 + (\sqrt{3}/2)^2 = 5/4

g(\pi/2)=1

g(5\pi/6) = 1/2 + (-\sqrt{3}/2)^2 = 5/4

, and

g(\pi)=1

. Comparing these values, the absolute maximum is 5/4.

Question 14

A function $f$ is twice-differentiable. At $x=c$ , we find that $f'(c) = 0$ and $f''(c) = 0$ . Which of the following conclusions is logically valid?

$f$ must have a point of inflection at $x=c$ .
$f$ has neither a local maximum nor a local minimum at $x=c$ .
$f$ must have either a local maximum or a local minimum at $x=c$ .
The Second Derivative Test is inconclusive regarding local extrema at $x=c$ . (correct answer)

Explanation: When

f'(c) = 0

and

f''(c) = 0

, the Second Derivative Test fails to provide information about whether

f

has a local extremum at

c

. For example, if

f(x)=x^4

, then

f'(0)=0

and

f''(0)=0

, and

f

has a local minimum at

x=0

. If

g(x)=x^3

, then

g'(0)=0

and

g''(0)=0

, and

g

has no extremum at

x=0

. Because different outcomes are possible, the only valid conclusion is that the test is inconclusive.

Question 15

What is the absolute minimum value of the function $f(x) = x^3 - 6x^2 + 5$ on the closed interval $[-3, 5]$ ?

$-76$ (correct answer)
$-27$
$-20$
$5$

Explanation: To find the absolute minimum value on a closed interval, we must test the critical points within the interval and the endpoints of the interval. First, find the derivative of

f(x)

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

Next, find the critical points by setting

f'(x)=0

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

The critical points are

x=0

and

x=4

. Both of these are within the interval

[-3, 5]

. Now, evaluate the function

f(x)

at the critical points and at the endpoints of the interval,

x=-3

and

x=5

f(-3) = (-3)^3 - 6(-3)^2 + 5 = -27 - 54 + 5 = -76

f(0) = (0)^3 - 6(0)^2 + 5 = 5

f(4) = (4)^3 - 6(4)^2 + 5 = 64 - 96 + 5 = -27

f(5) = (5)^3 - 6(5)^2 + 5 = 125 - 150 + 5 = -20

Comparing these values, the absolute minimum value is

-76

. Distractor (B) is the value at the local minimum, and (C) is the value at an endpoint. Distractor (D) is the value at the local maximum.

Question 16

For what values of the constant $k$ does the function $f(x) = x^3 - kx^2 + 3x$ have exactly one critical point?

$k=3$ or $k=-3$ (correct answer)
$k > 3$ or $k < -3$
$-3 < k < 3$
$k=0$ only

Explanation: Critical points occur where the derivative

f'(x)

is zero or undefined. Since

f(x)

is a polynomial, its derivative will be defined everywhere. First, find the derivative:

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

. To find the critical points, we set the derivative equal to zero:

3x^2 - 2kx + 3 = 0

. This is a quadratic equation in terms of

x

. The number of critical points is the number of real solutions to this equation. The function will have exactly one critical point if this quadratic equation has exactly one real solution. This occurs when the discriminant (

b^2 - 4ac

) is equal to zero. For the quadratic

3x^2 - 2kx + 3 = 0

, we have

a=3

b=-2k

, and

c=3

. The discriminant is:

(-2k)^2 - 4(3)(3) = 4k^2 - 36

. Set the discriminant to zero and solve for

k

4k^2 - 36 = 0

4k^2 = 36

k^2 = 9

k = \pm 3

Thus, the function has exactly one critical point when

k=3

k=-3

Question 17

Let $f$ be a function that is continuous on the closed interval $[a, b]$ . Which of the following statements is necessarily true?

If $f$ has an absolute maximum at $x=c$ in $(a,b)$ , then $f'(c)=0$ .
The absolute maximum value of $f$ must occur at a critical point.
$f$ must have at least one critical point in the interval $(a,b)$ .
If $f(a) = f(b)$ , then $f$ must have an absolute extremum at some point $c$ in $(a,b)$ . (correct answer)

Explanation: This question tests the implications of the Extreme Value Theorem (EVT). (A) is false. By Fermat's Theorem, if

f

has a local extremum at

c

and

f'(c)

exists, then

f'(c)=0

. However, the absolute maximum could occur where the derivative is undefined. For example,

f(x)=-|x|

[-1,1]

has an absolute maximum at

c=0

, but

f'(0)

is undefined. (B) is false. The absolute maximum of a function on a closed interval can occur at an endpoint (

x=a

x=b

), which is not required to be a critical point. For example,

f(x)=x

[0,1]

has its absolute maximum at

x=1

, which is not a critical point. (C) is false. A function can have its absolute maximum and minimum at the endpoints. For example,

f(x)=x

[0,1]

is continuous on a closed interval but has no critical points in

(0,1)

. (D) is true. By the EVT,

f

must attain an absolute maximum and minimum on

[a,b]

. If both the absolute maximum and minimum occur at the endpoints, then since

f(a)=f(b)

, the maximum and minimum values are the same, meaning the function must be constant. In this case, any point

c

(a,b)

is an absolute extremum. If the function is not constant, then at least one of its absolute extrema must have a value different from

f(a)=f(b)

and therefore must occur at a point

c

in the open interval

(a,b)

Question 18

What is the absolute maximum value of the function $f(x) = x e^{-x^2/8}$ on the interval $(0, \infty)$ ?

The function has no absolute maximum.
$0$
$\frac{2}{\sqrt{e}}$ (correct answer)
$\frac{-2}{\sqrt{e}}$

Explanation: The Extreme Value Theorem does not apply since the interval

(0, \infty)

is not closed. We must find critical points and analyze the function's behavior at the boundaries of the interval. First, find the derivative using the product rule:

f'(x) = (1)e^{-x^2/8} + x \cdot e^{-x^2/8} \cdot (-\frac{2x}{8}) = e^{-x^2/8}(1 - \frac{x^2}{4})

Set

f'(x)=0

to find critical points. Since

e^{-x^2/8}

is always positive, we solve:

1 - \frac{x^2}{4} = 0 \implies x^2 = 4 \implies x = \pm 2

The only critical point in the interval

(0, \infty)

x=2

. Now, evaluate the function at the critical point and check the limits at the interval boundaries: At the critical point:

f(2) = 2e^{-2^2/8} = 2e^{-4/8} = 2e^{-1/2} = \frac{2}{\sqrt{e}}

. At the boundaries:

\lim_{x \to 0^+} f(x) = 0 \cdot e^0 = 0

\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x}{e^{x^2/8}}

. This is an

\frac{\infty}{\infty}

form, so we use L'Hôpital's Rule:

\lim_{x \to \infty} \frac{1}{\frac{2x}{8}e^{x^2/8}} = \lim_{x \to \infty} \frac{4}{xe^{x^2/8}} = 0

. Since the function's value is 0 at both ends of the interval and is positive in between, the value at the only critical point,

f(2) = \frac{2}{\sqrt{e}}

, must be the absolute maximum.

Question 19

Let $f(x) = (x-2)^{2/3}(2x+1)$ . What is the sum of the x-coordinates of all critical points of $f$ ?

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

Explanation: To find the critical points of

f(x)

, we need to find its derivative,

f'(x)

, and identify where

f'(x)=0

f'(x)

is undefined. The domain of

f(x)

is all real numbers. Using the product rule and chain rule:

f'(x) = \frac{d}{dx}[(x-2)^{2/3}] \cdot (2x+1) + (x-2)^{2/3} \cdot \frac{d}{dx}[2x+1]

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

To simplify, find a common denominator:

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

f'(x) = \frac{4x+2 + 6(x-2)}{3(x-2)^{1/3}} = \frac{4x+2+6x-12}{3(x-2)^{1/3}} = \frac{10x-10}{3(x-2)^{1/3}}

Critical points occur where the numerator is zero or the denominator is zero.

$f'(x)=0$ : $10x-10=0 \implies x=1$ .
$f'(x)$ is undefined: $3(x-2)^{1/3}=0 \implies x=2$ .

Both

x=1

and

x=2

are in the domain of

f(x)

, so both are critical points. The sum of their x-coordinates is

1+2=3

Question 20

Let $f(x) = \sqrt{x^2 - 6x + 25}$ . What is the absolute minimum value of $f(x)$ ?

3
4 (correct answer)
5
16

Explanation: The minimum value of

f(x)

will occur when the quadratic expression inside the square root,

g(x) = x^2 - 6x + 25

, is at its minimum. Since

g(x)

is an upward-opening parabola, its minimum is at its vertex. The x-coordinate of the vertex is

x = -\frac{-6}{2(1)} = 3

. The minimum value of the quadratic is

g(3) = 3^2 - 6(3) + 25 = 9 - 18 + 25 = 16

. Therefore, the minimum value of

f(x)

\sqrt{16} = 4