Calculus 1 Quiz: Mean Value Theorem

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Mean Value TheoremQuestion 1 of 20

For the function $f(x) = x^3 - 3x$ on the interval $[-2, 2]$ , what is the sum of all values $c$ in the interval $(-2, 2)$ that satisfy the conclusion of the Mean Value Theorem?

0

\frac{2}{\sqrt{3}}

4

No such values exist.

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Calculus 1 Quiz

Calculus 1 Quiz: Mean Value Theorem

Practice Mean Value Theorem 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 Mean Value Theorem, 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

For the function $f(x) = x^3 - 3x$ on the interval $[-2, 2]$ , what is the sum of all values $c$ in the interval $(-2, 2)$ that satisfy the conclusion of the Mean Value Theorem?

$0$ (correct answer)
$\frac{2}{\sqrt{3}}$
$4$
No such values exist.

Explanation: First, calculate the slope of the secant line on

[-2, 2]

f(2) = 2^3 - 3(2) = 8 - 6 = 2

f(-2) = (-2)^3 - 3(-2) = -8 + 6 = -2

. The slope is

\frac{f(2) - f(-2)}{2 - (-2)} = \frac{2 - (-2)}{4} = \frac{4}{4} = 1

. Next, find the derivative:

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

. According to the Mean Value Theorem, we must find

c

such that

f'(c) = 1

. So,

3c^2 - 3 = 1

, which gives

3c^2 = 4

, so

c^2 = \frac{4}{3}

. The possible values for

c

are

c = \frac{2}{\sqrt{3}}

and

c = -\frac{2}{\sqrt{3}}

. Both of these values lie within the open interval

(-2, 2)

. The sum of these values is

\frac{2}{\sqrt{3}} + (-\frac{2}{\sqrt{3}}) = 0

Question 2

Let ƒ be a function that is continuous on [a, b] and differentiable on (a, b). If ƒ'(x) is never zero on (a, b), what is the maximum number of times the graph of ƒ can intersect the line y = ƒ(a)?

0
1 (correct answer)
2
Cannot be determined

Explanation: The graph of ƒ intersects the line y = ƒ(a) at x = a. Suppose it intersects the line at another point, x = d, where a < d ≤ b. This would mean ƒ(a) = ƒ(d). By Rolle's Theorem (a special case of MVT), there must exist a c in (a, d) such that ƒ'(c) = 0. However, the problem states that ƒ'(x) is never zero on (a, b). This is a contradiction. Therefore, the graph cannot intersect the line y = ƒ(a) at any point other than x = a. The maximum number of intersections is 1.

Question 3

A car passes a checkpoint at time $t=0$ hours and a second checkpoint 120 miles away at $t=1.5$ hours. The car's position function $s(t)$ is continuous and differentiable. The driver is issued a speeding ticket for exceeding the 65 mph speed limit. Which mathematical theorem is the primary justification that the driver must have been speeding at some instant?

The Intermediate Value Theorem
The Extreme Value Theorem
The Squeeze Theorem
The Mean Value Theorem (correct answer)

Explanation: The driver's average velocity over the trip is the total distance divided by the total time:

\frac{120 \text{ miles}}{1.5 \text{ hours}} = 80

mph. Let

s(t)

be the car's position at time

t

. The average velocity is

\frac{s(1.5) - s(0)}{1.5 - 0}

. The Mean Value Theorem states that if

s(t)

is continuous on

[0, 1.5]

and differentiable on

(0, 1.5)

, then there must be some time

c

(0, 1.5)

where the instantaneous velocity,

s'(c)

, is equal to the average velocity. Therefore, there was an instant

c

where the car's velocity was exactly 80 mph. Since 80 mph > 65 mph, the driver was speeding at that instant. This provides the justification for the ticket.

Question 4

Let $f(x) = x^3 - x + k$ for some constant $k$ . If the Mean Value Theorem is applied on the interval $[0, 3]$ , it is found that the value $c = \sqrt{3}$ satisfies the conclusion. What can be concluded about the value of the constant $k$ ?

$k = 0$
$k = 8$
$k$ can be any real number. (correct answer)
$k$ must be negative.

Explanation: The Mean Value Theorem states that for a continuous, differentiable function on an interval

[a,b]

, there exists some point

c

where

f'(c) = \frac{f(b) - f(a)}{b - a}

. The key insight here is that the conclusion of MVT depends only on the derivative and the endpoint values. Let's verify this works with

c = \sqrt{3}

. First, find

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

, so

f'(\sqrt{3}) = 3(3) - 1 = 8

. Next, calculate the average rate of change:

\frac{f(3) - f(0)}{3 - 0} = \frac{(27 - 3 + k) - (0 - 0 + k)}{3} = \frac{24 + k - k}{3} = 8

. Notice something crucial: the constant

k

cancels out completely! This happens because

f(3) - f(0) = (3^3 - 3 + k) - (0^3 - 0 + k) = 24 + k - k = 24

. Since

f'(\sqrt{3}) = 8

equals the average rate of change of

8

regardless of

k

's value, the Mean Value Theorem conclusion holds for any real number

k

. Answer choice (A)

k = 0

is incorrect because

k

doesn't need to be zero specifically. Choice (B)

k = 8

is wrong for the same reason—this particular value isn't required. Choice (D)

k

must be negative is false since we've shown

k

can be any real number, including positive values. Study tip: When applying MVT, remember that adding constants to functions doesn't affect the theorem's conclusion since constants disappear when computing differences

f(b) - f(a)

. The shape of the curve matters, not its vertical position.

Question 5

The position of a particle moving along a straight line is given by $p(t) = t^3 - 6t^2 + 12t - 8$ for $t \ge 0$ . According to the Mean Value Theorem, there are two times, $c_1$ and $c_2$ , in the interval $(1, 3)$ where the particle's instantaneous velocity equals its average velocity over $[1, 3]$ . What is the sum $c_1 + c_2$ ?

$2$
$2\sqrt{3}$
$4$ (correct answer)
$11/3$

Explanation: When you see a question combining position functions with the Mean Value Theorem, you're dealing with the relationship between average and instantaneous rates of change. The Mean Value Theorem guarantees that somewhere in an interval, the instantaneous rate equals the average rate. First, find the average velocity over

[1,3]

. Calculate

p(3) = 27 - 54 + 36 - 8 = 1

and

p(1) = 1 - 6 + 12 - 8 = -1

. So the average velocity is

\frac{p(3) - p(1)}{3 - 1} = \frac{1 - (-1)}{2} = 1

. Next, find when the instantaneous velocity equals this average. The velocity function is

v(t) = p'(t) = 3t^2 - 12t + 12

. Set this equal to the average velocity:

3t^2 - 12t + 12 = 1

, which simplifies to

3t^2 - 12t + 11 = 0

. Using the quadratic formula:

t = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3}

. Both values lie in

(1,3)

, so

c_1 + c_2 = (2 + \frac{\sqrt{3}}{3}) + (2 - \frac{\sqrt{3}}{3}) = 4

. Choice A (

2

) would be correct if you mistakenly used the average of the interval endpoints. Choice B (

2\sqrt{3}

) might result from misapplying the quadratic formula. Choice D (

11/3

) could come from algebraic errors when solving the quadratic equation. The correct answer is C. Study tip: When applying MVT to motion problems, always verify your critical points lie within the given interval, and remember that the sum of roots

\frac{b \pm \sqrt{...}}{2a}

equals

\frac{2b}{2a} = \frac{b}{a}

Question 6

A function $f$ is differentiable on $[0, 5]$ with $f(0) = 10$ . The derivative of the function is bounded such that $-2 \le f'(x) \le 3$ for all $x$ in $[0, 5]$ . Which of the following is not a possible value for $f(5)$ ?

$1$
$15$
$24$
$26$ (correct answer)

Explanation: By the Mean Value Theorem, there exists a

c

(0, 5)

such that

f'(c) = \frac{f(5) - f(0)}{5 - 0}

. This means

f(5) - f(0) = 5f'(c)

. Substituting

f(0)=10

, we get

f(5) = 10 + 5f'(c)

. We are given that

-2 \le f'(x) \le 3

for all

x

, so this applies to

f'(c)

. We can find the range of possible values for

f(5)

by substituting the minimum and maximum values of

f'(c)

. The minimum value is

10 + 5(-2) = 10 - 10 = 0

. The maximum value is

10 + 5(3) = 10 + 15 = 25

. Therefore, the value of

f(5)

must be in the interval

[0, 25]

. The only value among the choices that is outside this interval is 26.

Question 7

Consider the function $f(x) = \arcsin(\frac{x}{2})$ . The Mean Value Theorem applied to $f(x)$ on the interval $[-1, 1]$ guarantees the existence of a number $c$ in $(-1, 1)$ . What is the value of $c^2$ ?

$4(1 - \frac{9}{\pi^2})$ (correct answer)
$4(1 - \frac{36}{\pi^2})$
$2(1 - \frac{9}{\pi^2})$
$1 - \frac{3}{\pi}$

Explanation: First, find the average rate of change on

[-1, 1]

f(1) = \arcsin(1/2) = \pi/6

f(-1) = \arcsin(-1/2) = -\pi/6

. The slope of the secant line is

\frac{\pi/6 - (-\pi/6)}{1 - (-1)} = \frac{2\pi/6}{2} = \frac{\pi}{6}

. Next, find the derivative using the chain rule:

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

. Set

f'(c) = \pi/6

\frac{1}{2\sqrt{1 - c^2/4}} = \frac{\pi}{6}

. This simplifies to

6 = 2\pi\sqrt{1 - c^2/4}

, or

3/\pi = \sqrt{1 - c^2/4}

. Squaring both sides gives

\frac{9}{\pi^2} = 1 - \frac{c^2}{4}

. Rearranging terms, we get

\frac{c^2}{4} = 1 - \frac{9}{\pi^2}

, so

c^2 = 4(1 - \frac{9}{\pi^2})

Question 8

Let ƒ be a function that is differentiable on the set of all real numbers. If ƒ(2) = 5 and ƒ'(x) ≥ 3 for all x, what is the smallest possible value of ƒ(6)?

12
15
17 (correct answer)
18

Explanation: By the Mean Value Theorem on the interval [2, 6], there exists a number c in (2, 6) such that ƒ'(c) = (ƒ(6) - ƒ(2)) / (6 - 2). We are given that ƒ'(x) ≥ 3 for all x, so ƒ'(c) ≥ 3. Substituting the known values, we get (ƒ(6) - 5) / 4 ≥ 3. Multiplying by 4 gives ƒ(6) - 5 ≥ 12. Adding 5 to both sides gives ƒ(6) ≥ 17. Therefore, the smallest possible value of ƒ(6) is 17.

Question 9

Let ƒ(x) = x³ - 6x² + 5x. On the interval [0, 4], the Mean Value Theorem guarantees the existence of at least one value c such that ƒ'(c) equals the average rate of change on the interval. What is the sum of all such values of c?

2
4 (correct answer)
2 + (2√3)/3
8

Explanation: First, find the average rate of change. ƒ(0) = 0. ƒ(4) = 4³ - 6(4)² + 5(4) = 64 - 96 + 20 = -12. The average rate of change is (ƒ(4) - ƒ(0)) / (4 - 0) = (-12 - 0) / 4 = -3. Next, find the derivative: ƒ'(x) = 3x² - 12x + 5. Set ƒ'(c) = -3: 3c² - 12c + 5 = -3, which simplifies to 3c² - 12c + 8 = 0. Using the quadratic formula, c = (12 ± √(144 - 438)) / 6 = (12 ± √48) / 6 = 2 ± (4√3)/6 = 2 ± (2√3)/3. Both c₁ = 2 - (2√3)/3 ≈ 0.845 and c₂ = 2 + (2√3)/3 ≈ 3.155 are in the interval (0, 4). The sum of these values is (2 - (2√3)/3) + (2 + (2√3)/3) = 4. Alternatively, by Vieta's formulas, the sum of the roots of 3c² - 12c + 8 = 0 is -(-12)/3 = 4.

Question 10

A function g(x) is continuous on [1, 8] and differentiable on (1, 8). If g(1) = 10 and g'(x) ≤ -2 for all x in (1, 8), what is the maximum possible value of g(8)?

-14
-6
-4 (correct answer)
24

Explanation: According to the Mean Value Theorem, there exists c in (1, 8) such that g'(c) = (g(8) - g(1)) / (8 - 1). We know g'(c) ≤ -2 and g(1) = 10. So, (g(8) - 10) / 7 ≤ -2. Multiplying by 7, we get g(8) - 10 ≤ -14. Adding 10 to both sides, we find g(8) ≤ -4. Thus, the maximum possible value of g(8) is -4.

Question 11

Let ƒ be a twice-differentiable function. Suppose ƒ(1) = 3, ƒ(3) = 7, and ƒ(5) = 15. Which of the following statements must be true?

There exists a c in (1, 5) such that ƒ'(c) = 3.
There exists a c in (1, 5) such that ƒ(c) = 10.
There exists a c in (1, 5) such that ƒ''(c) < 0.
There exists a c in (1, 5) such that ƒ''(c) > 0. (correct answer)

Explanation: First, apply the Mean Value Theorem to ƒ on [1, 3]. There exists c₁ in (1, 3) such that ƒ'(c₁) = (ƒ(3) - ƒ(1)) / (3 - 1) = (7 - 3) / 2 = 2. Next, apply the MVT to ƒ on [3, 5]. There exists c₂ in (3, 5) such that ƒ'(c₂) = (ƒ(5) - ƒ(3)) / (5 - 3) = (15 - 7) / 2 = 4. Now, apply the MVT to the function ƒ' on the interval [c₁, c₂]. Since ƒ is twice-differentiable, ƒ' is differentiable. There exists d in (c₁, c₂) such that ƒ''(d) = (ƒ'(c₂) - ƒ'(c₁)) / (c₂ - c₁). Substituting the values, ƒ''(d) = (4 - 2) / (c₂ - c₁) = 2 / (c₂ - c₁). Since c₂ > c₁, the denominator is positive, so ƒ''(d) > 0. As (c₁, c₂) is a subinterval of (1, 5), d is in (1, 5).

Question 12

Let ƒ be a differentiable function on the set of all real numbers. If the equation ƒ'(x) = 0 has exactly two distinct solutions, what is the maximum number of distinct solutions the equation ƒ(x) = k can have for any real constant k?

1
2
3 (correct answer)
4

Explanation: This is an application of Rolle's Theorem, a special case of the MVT. Rolle's Theorem states that between any two roots of a differentiable function, there must be at least one point where the derivative is zero. Suppose ƒ(x) = k has n distinct solutions x₁, x₂, ..., xₙ. These are roots of the function g(x) = ƒ(x) - k. Between each adjacent pair of roots, there must be a point where g'(x) = ƒ'(x) = 0. If there are n roots, there are n-1 intervals between them, implying at least n-1 points where ƒ'(x) = 0. We are given that ƒ'(x) = 0 has exactly two solutions. Therefore, n-1 ≤ 2, which implies n ≤ 3. The maximum number of distinct solutions is 3.

Question 13

Let ƒ be a function that is continuous on [a, b] and differentiable on (a, b). If for every subinterval [x₁, x₂] ⊆ [a, b], the value c guaranteed by the Mean Value Theorem is always the midpoint of the subinterval, c = (x₁ + x₂) / 2, what type of function must ƒ(x) be?

A linear function
A quadratic function (correct answer)
A constant function
A cubic function

Explanation: The condition is (ƒ(x₂) - ƒ(x₁)) / (x₂ - x₁) = ƒ'((x₁ + x₂)/2). Let's test a general quadratic function, ƒ(x) = Ax² + Bx + C. The derivative is ƒ'(x) = 2Ax + B. The average rate of change is [A(x₂² - x₁²) + B(x₂ - x₁)] / (x₂ - x₁) = A(x₂ + x₁) + B. The instantaneous rate of change at the midpoint is ƒ'((x₁ + x₂)/2) = 2A((x₁ + x₂)/2) + B = A(x₁ + x₂) + B. The expressions match for any quadratic function. This property does not hold for cubic or higher-degree polynomials, and while it holds for linear functions (a subset of quadratics), 'quadratic' is the most general correct description.

Question 14

If a function ƒ is differentiable for all x and ƒ(x₁) = ƒ(x₂) for two distinct points x₁ and x₂, Rolle's Theorem guarantees a point c between x₁ and x₂ such that ƒ'(c) = 0. If ƒ'(x) = 4x³ - 12x, what is the maximum possible number of distinct real roots of ƒ(x)?

2
3
4 (correct answer)
5

Explanation: We first find the roots of ƒ'(x) = 0. We have 4x³ - 12x = 4x(x² - 3) = 0, which has three distinct roots: x = 0, x = √3, and x = -√3. By Rolle's Theorem, between any two roots of ƒ(x), there must be a root of ƒ'(x). If ƒ(x) had n distinct roots, then ƒ'(x) must have at least n-1 distinct roots. Since ƒ'(x) has exactly 3 roots, we must have n-1 ≤ 3, which implies n ≤ 4. Therefore, ƒ(x) can have at most 4 distinct real roots.

Question 15

Let ƒ be a differentiable function on [a, b]. The expression (ƒ(b) - ƒ(a)) / (b - a) represents the slope of the secant line connecting (a, ƒ(a)) and (b, ƒ(b)). The Mean Value Theorem guarantees a point c in (a, b) where ƒ'(c) equals this value. For ƒ(x) = x³ on [-1, 2], what is the value of c?

1 (correct answer)
1/2
√3
√2

Explanation: First, calculate the average rate of change for ƒ(x) = x³ on [-1, 2]. ƒ(2) = 2³ = 8. ƒ(-1) = (-1)³ = -1. The average rate of change is (8 - (-1)) / (2 - (-1)) = 9 / 3 = 3. Next, find the derivative: ƒ'(x) = 3x². Set ƒ'(c) = 3, which gives 3c² = 3, so c² = 1. The solutions are c = 1 and c = -1. We must choose the value of c that lies in the open interval (-1, 2). Therefore, c = 1.

Question 16

Consider the function g(x) = x + cos(x) on the interval [0, π/2]. Let c be the value guaranteed by the Mean Value Theorem. Which of the following statements is true about c?

sin(c) = 2/π (correct answer)
c = π/6
c = π/4
cos(c) = 2/π

Explanation: When you encounter a Mean Value Theorem problem, you need to apply the theorem systematically and then solve for the specific value it guarantees. The Mean Value Theorem states that for a continuous, differentiable function on

[a,b]

, there exists some

c

(a,b)

where

f'(c) = \frac{f(b) - f(a)}{b - a}

. For

g(x) = x + \cos(x)

[0, \pi/2]

, let's find this relationship. First,

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

. Next, we calculate the average rate of change:

\frac{g(\pi/2) - g(0)}{\pi/2 - 0} = \frac{(\pi/2 + 0) - (0 + 1)}{\pi/2} = \frac{\pi/2 - 1}{\pi/2} = 1 - \frac{2}{\pi}

. Setting

g'(c) =

average rate of change:

1 - \sin(c) = 1 - \frac{2}{\pi}

. Solving:

-\sin(c) = -\frac{2}{\pi}

, so

\sin(c) = \frac{2}{\pi}

. Looking at the choices: A)

\sin(c) = \frac{2}{\pi}

matches our result exactly. B)

c = \pi/6

would give

\sin(c) = 1/2

, which doesn't equal

2/\pi \approx 0.637

. C)

c = \pi/4

would give

\sin(c) = \sqrt{2}/2 \approx 0.707

, also incorrect. D)

\cos(c) = 2/\pi

confuses cosine with sine. Study tip: In Mean Value Theorem problems, always set up the equation

f'(c) = \frac{f(b) - f(a)}{b - a}

first, then solve algebraically. Don't try to guess the value of

c

directly—focus on what the theorem tells you about

f'(c)

Question 17

The Mean Value Theorem does not apply to ƒ(x) = |x² - 4| on the interval [-3, 3]. Which hypothesis of the theorem is not satisfied?

ƒ(x) is not differentiable on (-3, 3). (correct answer)
ƒ(x) is not defined at all points in [-3, 3].
ƒ(x) is not continuous on [-3, 3].
ƒ(-3) is not equal to ƒ(3).

Explanation: The Mean Value Theorem is a fundamental result in calculus that requires two key conditions: the function must be continuous on the closed interval and differentiable on the open interval. When you encounter questions about MVT applications, always check both conditions systematically. For

f(x) = |x² - 4|

[-3, 3]

, you need to analyze where the absolute value creates problems. First, determine where

x² - 4 = 0

, which occurs at

x = ±2

. At these points, the expression inside the absolute value changes sign, so

f(x) = |x² - 4|

becomes:

$f(x) = x² - 4$ when $x² - 4 ≥ 0$ (outside $[-2, 2]$ )
$f(x) = -(x² - 4) = 4 - x²$ when $x² - 4 < 0$ (inside $(-2, 2)$ )

The function is continuous everywhere since absolute value functions are continuous. However, at

x = ±2

, the derivative changes abruptly from

2x

-2x

, creating sharp corners where the function isn't differentiable. Choice A is correct because differentiability fails at

x = ±2

within the open interval

(-3, 3)

. Choice B is wrong because

f(x)

is clearly defined everywhere on

[-3, 3]

. Choice C is incorrect since absolute value functions maintain continuity. Choice D is irrelevant because MVT doesn't require the endpoint values to be equal—that's Rolle's Theorem. Study tip: When checking MVT applicability, always look for sharp corners or cusps in absolute value functions by finding where the inside expression equals zero.

Question 18

Let ƒ(x) = ln(x). On the interval [1, e²], the Mean Value Theorem guarantees the existence of a point c where the tangent line to the graph of ƒ is parallel to the secant line through the interval's endpoints. What is the value of c?

(e² - 1) / 2 (correct answer)
e² - 1
e
2 / (e² - 1)

Explanation: The Mean Value Theorem states that for a continuous, differentiable function on a closed interval, there exists at least one point where the instantaneous rate of change (derivative) equals the average rate of change over the interval. To find this point c, you need to set up the equation

f'(c) = \frac{f(b) - f(a)}{b - a}

. For

f(x) = \ln(x)

[1, e^2]

, first calculate the average rate of change:

\frac{\ln(e^2) - \ln(1)}{e^2 - 1} = \frac{2 - 0}{e^2 - 1} = \frac{2}{e^2 - 1}

. Since

f'(x) = \frac{1}{x}

, you need to solve

\frac{1}{c} = \frac{2}{e^2 - 1}

. Cross-multiplying gives

e^2 - 1 = 2c

, so

c = \frac{e^2 - 1}{2}

. Choice A is correct:

\frac{e^2 - 1}{2}

. Choice B (

e^2 - 1

) is twice the correct answer—this would result from forgetting to divide by 2 when solving for c. Choice C (

e

) might seem tempting as the midpoint of the interval in some sense, but it doesn't satisfy the MVT equation. Choice D (

\frac{2}{e^2 - 1}

) is actually the slope value itself, not the x-coordinate where this slope occurs. Remember: the Mean Value Theorem asks where the derivative equals the average rate of change. Always calculate the average rate first, then set it equal to the derivative and solve for the unknown point.

Question 19

The Mean Value Theorem requires a function ƒ to be continuous on [a, b] and differentiable on (a, b). Which of the following function-interval pairs demonstrates that the differentiability condition is necessary for the theorem's conclusion to hold?

ƒ(x) = x² on [0, 2]
ƒ(x) = |x| on [-1, 1] (correct answer)
ƒ(x) = 1/x on [-1, 1]
ƒ(x) = 5 on [0, 10]

Explanation: We need a function that is continuous on the closed interval but not differentiable on the open interval, and for which the conclusion of the MVT fails. For ƒ(x) = |x| on [-1, 1], the function is continuous. However, it is not differentiable at x=0. The average rate of change is (ƒ(1) - ƒ(-1)) / (1 - (-1)) = (1 - 1) / 2 = 0. The derivative is ƒ'(x) = -1 for x < 0 and ƒ'(x) = 1 for x > 0. There is no c in (-1, 1) where ƒ'(c) = 0. Thus, the conclusion fails, demonstrating the necessity of the differentiability hypothesis. Choice C fails the continuity condition. Choices A and D satisfy all hypotheses.

Question 20

Let ƒ be a function that is continuous on [0, 4] and differentiable on (0, 4). If ƒ(0) = 1 and ƒ(4) = 9, the Mean Value Theorem guarantees a point c in (0, 4) such that ƒ'(c) = 2. Which additional condition would guarantee that there is a point d in (0, 4) such that ƒ(d) = 5?

ƒ'(x) > 0 for all x in (0, 4).
ƒ is concave up on (0, 4).
ƒ is also differentiable at x = 0 and x = 4.
No additional condition is needed; this is already guaranteed. (correct answer)

Explanation: The existence of a point d in (0, 4) such that ƒ(d) = 5 is guaranteed by the Intermediate Value Theorem (IVT), not the Mean Value Theorem. The IVT applies because ƒ is continuous on [0, 4], ƒ(0) = 1, and ƒ(4) = 9. Since 5 is a value between 1 and 9, there must be a point d in (0, 4) where ƒ(d) = 5. The information about the MVT is extra information that is not needed to answer the question. No additional conditions are required.