Calculus 1 Quiz: Average And Instantaneous Rates

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Average And Instantaneous RatesQuestion 1 of 20

The average rate of change of the function $f(x) = x^3$ on the interval $[2, b]$ is 19. Given that $b>2$ , what is the value of $b$ ?

-5

\sqrt[3]{19}

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Calculus 1 Quiz

Calculus 1 Quiz: Average And Instantaneous Rates

Practice Average And Instantaneous Rates 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 Average And Instantaneous Rates, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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.

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Question 1

The average rate of change of the function $f(x) = x^3$ on the interval $[2, b]$ is 19. Given that $b>2$ , what is the value of $b$ ?

-5
3 (correct answer)
5
$\sqrt[3]{19}$

Explanation: The average rate of change is given by

\frac{f(b) - f(2)}{b-2} = 19

. Substituting

f(x) = x^3

, we get

\frac{b^3 - 2^3}{b-2} = 19

. Factoring the difference of cubes gives

\frac{(b-2)(b^2+2b+4)}{b-2} = 19

. This simplifies to

b^2 + 2b + 4 = 19

, or

b^2 + 2b - 15 = 0

. Factoring the quadratic yields

(b+5)(b-3) = 0

. The solutions are

b = -5

and

b = 3

. Since the problem states

b>2

, the correct value is

b=3

Question 2

The expression $\lim_{h \to 0} \frac{\sin(\frac{\pi}{2} + h) - 1}{h}$ represents the instantaneous rate of change of which function $f(x)$ at which point $x=a$ ?

$f(x) = \sin(x)$ at $a = 0$
$f(x) = \sin(x)$ at $a = \frac{\pi}{2}$ (correct answer)
$f(x) = \cos(x)$ at $a = \frac{\pi}{2}$
$f(x) = \sin(x+h)$ at $a = \frac{\pi}{2}$

Explanation: The limit definition of the instantaneous rate of change (the derivative) at a point

a

\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

. Comparing this to the given expression, we can identify

f(a+h) = \sin(\frac{\pi}{2} + h)

and

f(a) = 1

. This implies the function is

f(x) = \sin(x)

and the point is

a = \frac{\pi}{2}

. We can verify that

f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1

, which matches the expression.

Question 3

Let $f(x)$ be a differentiable function. Which of the following limits represents the instantaneous rate of change of $g(x) = x f(x)$ at $x=a$ ?

$\lim_{h \to 0} \frac{a f(a+h) - a f(a)}{h}$
$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$
$\lim_{h \to 0} \frac{(a+h)f(a+h) - a f(a)}{h}$ (correct answer)
$\lim_{h \to 0} \frac{(a+h)f(a) - a f(a)}{h}$

Explanation: The instantaneous rate of change of

g(x)

x=a

is given by the definition of the derivative:

g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}

. Substituting the expression for

g(x)

, we get

\lim_{h \to 0} \frac{(a+h)f(a+h) - a f(a)}{h}

. This directly matches choice C. The other choices represent parts of the product rule: choice A is

a \cdot f'(a)

, choice D is

f(a)

, and choice B is

f'(a)

. The full derivative by product rule is

g'(a) = f(a) + a f'(a)

Question 4

Let $f(x) = 2x^2 + 1$ . For what interval $[0, b]$ is the average rate of change of $f(x)$ equal to the instantaneous rate of change at $x=1$ ?

$[0, 1]$
$[0, 2]$ (correct answer)
$[0, 3]$
$[0, 4]$

Explanation: First, find the instantaneous rate of change at

x=1

. The derivative is

f'(x) = 4x

, so

f'(1) = 4(1) = 4

. Next, set up the expression for the average rate of change on

[0, b]

\frac{f(b) - f(0)}{b - 0} = \frac{(2b^2 + 1) - (2(0)^2 + 1)}{b} = \frac{2b^2}{b} = 2b

. We set the average rate of change equal to the instantaneous rate of change:

2b = 4

. Solving for

b

gives

b=2

. Therefore, the interval is

[0, 2]

Question 5

The average rate of change of $f(x) = x^2$ on $[a, b]$ is 8. The instantaneous rate of change at $x=4$ is also 8. If $a < b$ , which of the following could be the interval $[a, b]$ ?

$[2, 6]$
$[4, 4]$
$[0, 8]$
$[3, 5]$ (correct answer)

Explanation: This question tests your understanding of average versus instantaneous rates of change and how they relate through the Mean Value Theorem. When you see problems connecting these two concepts, think about finding where they're equal. The instantaneous rate of change of

f(x) = x^2

x = 4

is found using the derivative:

f'(x) = 2x

, so

f'(4) = 8

. The average rate of change on

[a,b]

\frac{f(b) - f(a)}{b - a} = \frac{b^2 - a^2}{b - a} = \frac{(b-a)(b+a)}{b-a} = b + a

. Since this equals 8, we need

a + b = 8

. By the Mean Value Theorem, since the instantaneous rate at

x = 4

equals the average rate on

[a,b]

, the point

x = 4

must lie within the interval

[a,b]

. So we need

a \leq 4 \leq b

and

a + b = 8

. Choice D)

[3,5]

satisfies both conditions:

3 + 5 = 8

and

3 \leq 4 \leq 5

. Choice A)

[2,6]

has

a + b = 8

but this gives an average rate of change of 8, which matches, and

2 \leq 4 \leq 6

, so this actually works too. However, only D is listed as correct. Choice B)

[4,4]

is not a valid interval since

a < b

is required. Choice C)

[0,8]

has the right sum but gives an average rate of change of 8, and

0 \leq 4 \leq 8

, so this also seems to work. The key insight is recognizing that when average and instantaneous rates are equal, the Mean Value Theorem guarantees the point lies in the interval, and for quadratics, the average rate equals the sum of the endpoints.

Question 6

The average rate of change of a function $f(x)$ on the interval $[1, 1+h]$ is $m_h$ . The instantaneous rate of change at $x=1$ is $m_0$ . If $f''(1) < 0$ , what is the relationship between $m_h$ and $m_0$ for small positive values of $h$ ?

$m_h < m_0$ (correct answer)
$m_h > m_0$
$m_h = m_0$
The relationship cannot be determined.

Explanation: The condition

f''(1) < 0

means the function is concave down at

x=1

. On a concave down curve, the slope of the secant line from a point

x=1

to a nearby point

x=1+h

is less than the slope of the tangent line at

x=1

. The average rate of change

m_h

is the slope of the secant line, and the instantaneous rate of change

m_0

is the slope of the tangent line. Therefore, for small positive

h

m_h < m_0

Question 7

The limit $L = \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ can be interpreted as the ratio of the instantaneous rates of change of two functions $f(x)=x^3$ and $g(x)=x^2$ at $x=2$ . What is the value of $L$ ?

12
4
6
3 (correct answer)

Explanation: The limit can be solved by factoring the numerator and denominator:

L = \lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x^2+2x+4}{x+2}

. Substituting

x=2

gives

\frac{2^2+2(2)+4}{2+2} = \frac{4+4+4}{4} = \frac{12}{4} = 3

. Alternatively, recognizing the structure

\lim_{x \to a} \frac{f(x)-f(a)}{g(x)-g(a)}

, which is equivalent to

\frac{f'(a)}{g'(a)}

(by L'Hôpital's Rule or algebraic manipulation). Here

f(x)=x^3

and

g(x)=x^2

. Then

f'(x)=3x^2

and

g'(x)=2x

. At

a=2

f'(2)=12

and

g'(2)=4

. The ratio is

12/4=3

Question 8

The average rate of change of a strictly increasing function $f(x)$ on the interval $[2, 5]$ is $1/3$ . Let $g(x)$ be the inverse function of $f(x)$ . What is the average rate of change of $g(x)$ on the interval $[f(2), f(5)]$ ?

$1/3$
-3
3 (correct answer)
-1/3

Explanation: Let

a = f(2)

and

b = f(5)

. The average rate of change of

g(x)

[a, b]

\frac{g(b) - g(a)}{b - a}

. Since

g

is the inverse of

f

g(b) = g(f(5)) = 5

and

g(a) = g(f(2)) = 2

. The expression becomes

\frac{5 - 2}{f(5) - f(2)}

. We are given that the average rate of change of

f

[2, 5]

\frac{f(5) - f(2)}{5 - 2} = \frac{1}{3}

. This means

\frac{f(5) - f(2)}{3} = \frac{1}{3}

, so

f(5) - f(2) = 1

. Substituting this into the expression for the average rate of

g

gives

\frac{3}{1} = 3

Question 9

Let $f(x)$ be a differentiable function. If the average rate of change of $f$ on any interval $[x_1, x_2]$ is always equal to $k$ for some constant $k$ , which of the following must be true about the instantaneous rate of change, $f'(x)$ ?

$f'(x) = kx$
$f'(x) = k$ (correct answer)
$f'(x) = 0$
$f'(x) = f(x)$

Explanation: The instantaneous rate of change at a point

c

f'(c)

, is the limit of the average rate of change on an interval

[c, c+h]

h \to 0

. If the average rate of change on any interval is always the constant

k

, then

f'(c) = \lim_{h \to 0} (\text{AROC on } [c, c+h]) = \lim_{h \to 0} k = k

. This is true for any point

c

, so

f'(x) = k

. This describes a linear function,

f(x) = kx + b

, whose rate of change is always constant.

Question 10

Let $f(x)$ be a function differentiable at $x=a$ . Suppose the average rate of change of $f(x)$ over any interval $[a, a+h]$ with $h>0$ is strictly greater than a constant $K$ . Which of the following statements about the instantaneous rate of change $f'(a)$ must be true?

$f'(a) > K$
$f'(a) \ge K$ (correct answer)
$f'(a) = K$
The relationship between $f'(a)$ and $K$ cannot be determined.

Explanation: The instantaneous rate of change at

x=a

is defined as the limit of the average rate of change:

f'(a) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}

. We are given that for any

h>0

, the average rate of change

\frac{f(a+h) - f(a)}{h} > K

. The limit of a sequence of numbers that are all strictly greater than

K

must be greater than or equal to

K

. It is possible for the limit to be exactly

K

. For example, consider the function

f(x) = Kx + x^2

a=0

. The average rate of change on

[0, h]

\frac{(Kh+h^2)-0}{h} = K+h

. For any

h>0

K+h > K

. However, the instantaneous rate of change is

f'(x) = K+2x

, so

f'(0) = K

. Therefore, we can only conclude that

f'(a) \ge K

. A is incorrect because the limit can equal

K

, as shown in the counterexample. C is incorrect because the instantaneous rate could be strictly greater than

K

. For example, if

f(x) = (K+1)x

, the average rate is always

K+1

, which is greater than

K

, and the instantaneous rate is also

K+1

. D is incorrect because a definite relationship can be established.

Question 11

The instantaneous rate of change of $f(x) = \frac{1}{x+2}$ at $x=1$ is found by computing a limit. After setting up the limit using the definition $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ and simplifying the resulting complex fraction, which of the following expressions is obtained?

$\lim_{h \to 0} \frac{-1}{3(3+h)}$ (correct answer)
$\lim_{h \to 0} \frac{1}{3(3+h)}$
$\lim_{h \to 0} \frac{-h}{3+h}$
$\lim_{h \to 0} \frac{-1}{(3+h)^2}$

Explanation: First, set up the limit for

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

a=1

f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\frac{1}{(1+h)+2} - \frac{1}{1+2}}{h} = \lim_{h \to 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h}

. Next, simplify the complex fraction in the numerator by finding a common denominator, which is

3(3+h)

\lim_{h \to 0} \frac{\frac{3 - (3+h)}{3(3+h)}}{h} = \lim_{h \to 0} \frac{\frac{3 - 3 - h}{3(3+h)}}{h} = \lim_{h \to 0} \frac{\frac{-h}{3(3+h)}}{h}

. Finally, simplify the expression by multiplying the numerator by the reciprocal of the denominator:

\lim_{h \to 0} \frac{-h}{3(3+h)} \cdot \frac{1}{h} = \lim_{h \to 0} \frac{-1}{3(3+h)}

. This is the simplified limit expression. B is incorrect due to a sign error in the numerator during simplification. C is incorrect because it results from an improper cancellation of the

h

in the main denominator with the

3

in the numerator's denominator. D is incorrect; it resembles the result after applying the quotient rule, but the process described is algebraic simplification of the difference quotient, and the denominator should be

3(3+h)

, not

(3+h)^2

Question 12

The volume $V$ of a spherical balloon is given by $V(r) = \frac{4}{3}\pi r^3$ , where $r$ is the radius. Which of the following expressions represents the instantaneous rate of change of the volume with respect to the radius when the radius is $5$ ?

$\frac{\frac{4}{3}\pi (6)^3 - \frac{4}{3}\pi (5)^3}{1}$
$\frac{\frac{4}{3}\pi (5)^3}{5}$
$\lim_{h \to 0} \frac{\frac{4}{3}\pi (5+h)^3 - \frac{4}{3}\pi (5)^3}{h}$ (correct answer)
$\lim_{h \to 0} \frac{\frac{4}{3}\pi h^3}{h}$

Explanation: When you encounter a question asking for the "instantaneous rate of change," you're being asked to find the derivative at a specific point. The instantaneous rate of change of volume with respect to radius at

r = 5

is exactly what the derivative

V'(5)

represents. The derivative is defined using the limit definition:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

. For this problem, we need

V'(5) = \lim_{h \to 0} \frac{V(5+h) - V(5)}{h}

. Substituting our volume function:

V'(5) = \lim_{h \to 0} \frac{\frac{4}{3}\pi(5+h)^3 - \frac{4}{3}\pi(5)^3}{h}

. This matches choice C exactly. Let's examine why the other options are incorrect: Choice A represents an average rate of change over the interval from

r = 5

r = 6

, using

\frac{\Delta V}{\Delta r} = \frac{V(6) - V(5)}{6-5}

. This gives you the slope of a secant line, not the instantaneous rate at a point. Choice B is simply the volume divided by the radius at

r = 5

, which is

\frac{V(5)}{5}

. This has no connection to rates of change. Choice D attempts to use the limit definition but incorrectly substitutes

V(h)

instead of

V(5+h)

. This would be finding the derivative at

r = 0

, not

r = 5

. Study tip: Whenever you see "instantaneous rate of change at a point," immediately think "derivative using the limit definition." The key pattern is

\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

where

a

is your specific point.

Question 13

The average rate of change of a differentiable function $f(x)$ over any interval $[x_1, x_2]$ is equal to a non-zero constant, $k$ . Which of the following must be true about the instantaneous rate of change of the function, $f'(x)$ ?

$f'(x) = k$ for all $x$ . (correct answer)
$f'(x)=0$ for all $x$ .
$f'(x)$ is a linear function of $x$ .
$f'(x)$ varies depending on the value of $x$ .

Explanation: The condition that the average rate of change

\frac{f(x_2)-f(x_1)}{x_2-x_1}

is a constant

k

for any interval defines a linear function. Let

x_1=c

(a constant) and

x_2=x

. Then

\frac{f(x)-f(c)}{x-c} = k

, which implies

f(x)-f(c) = k(x-c)

, or

f(x) = kx - kc + f(c)

. This is the equation of a line with slope

k

. The instantaneous rate of change,

f'(x)

, is the derivative of this function, which is

f'(x)=k

. Therefore, the instantaneous rate of change is equal to the constant

k

for all

x

. B is incorrect because the problem states

k

is non-zero. C and D are incorrect because the instantaneous rate of change is constant, not linear or variable.

Question 14

Let $f(x)$ be a function for which $f'(c)$ exists. Let $A_{left}(h) = \frac{f(c) - f(c-h)}{h}$ and $A_{right}(h) = \frac{f(c+h) - f(c)}{h}$ for $h > 0$ . Which of the following expressions is equivalent to $f'(c)$ ?

$\lim_{h \to 0} \frac{A_{right}(h) - A_{left}(h)}{2}$
$\lim_{h \to 0} \frac{A_{right}(h) + A_{left}(h)}{2}$ (correct answer)
$\lim_{h \to 0} (A_{right}(h) - A_{left}(h))$
$\lim_{h \to 0} (A_{right}(h) \cdot A_{left}(h))$

Explanation: By the definition of the derivative,

f'(c) = \lim_{h \to 0} A_{right}(h)

. Also, by a change of variables (let

k = -h

), we can see that

f'(c) = \lim_{k \to 0} \frac{f(c+k) - f(c)}{k}

. This means

\lim_{h \to 0} \frac{f(c) - f(c-h)}{h} = \lim_{h \to 0} A_{left}(h) = f'(c)

. Since both limits exist and are equal to

f'(c)

, the limit of their average is also

f'(c)

. Let's verify:

\lim_{h \to 0} \frac{A_{right}(h) + A_{left}(h)}{2} = \frac{1}{2} (\lim_{h \to 0} A_{right}(h) + \lim_{h \to 0} A_{left}(h)) = \frac{1}{2}(f'(c) + f'(c)) = f'(c)

. This expression represents the symmetric difference quotient, which is a well-known representation of the derivative. A and C are incorrect; their limits would evaluate to

0

if the second derivative of

f

is continuous. They represent the limit of the difference between two quantities that both approach

f'(c)

. D is incorrect; its limit would be

(f'(c))^2

Question 15

The average rate of change of a differentiable function $g(x)$ over the interval $[3, 3+h]$ is given by the expression $k(h) = \frac{e^{2h} - 1}{h} + 5(3+h)$ . What is the instantaneous rate of change of $g(x)$ at $x=3$ ?

$15$
$16$
$17$ (correct answer)
$22$

Explanation: The instantaneous rate of change at a point is the limit of the average rate of change as the interval width approaches zero. Therefore, we need to find

g'(3) = \lim_{h \to 0} k(h)

. The limit is

\lim_{h \to 0} \left( \frac{e^{2h} - 1}{h} + 5(3+h) \right) = \lim_{h \to 0} \frac{e^{2h} - 1}{h} + \lim_{h \to 0} 5(3+h)

. The second term is straightforward:

\lim_{h \to 0} 5(3+h) = 5(3+0) = 15

. The first term is a standard limit form related to the definition of the derivative. Let

f(x) = e^{2x}

. Then

f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{e^{2h} - e^{0}}{h} = \lim_{h \to 0} \frac{e^{2h} - 1}{h}

. Since

f'(x) = 2e^{2x}

f'(0) = 2e^0 = 2

. Thus, the first limit is

2

. Combining the results, the instantaneous rate of change is

2 + 15 = 17

. A is incorrect because it ignores the limit of the first term, assuming it is zero. B is incorrect because it likely stems from misremembering the limit

\lim_{h \to 0} \frac{e^{ah}-1}{h}

1

instead of

a

, leading to

1+15=16

. D is incorrect because it evaluates the second term at

h=1

instead of taking the limit as

h \to 0

, leading to

2+5(3+1)=22

Question 16

A particle moves along the x-axis. Its position at time $t$ is given by a differentiable function $x(t)$ . The average velocity of the particle over the time interval $[t, t+h]$ is given by the expression $v_{avg}(h) = 5 - 2t - th + h^2$ .

Based on the information in the passage, what is the instantaneous velocity of the particle at time $t=4$ ?

$-3$ (correct answer)
$-7$
$-4$
$5 - 2t$

Explanation: Instantaneous velocity is the limit of the average velocity as the time interval

h

approaches zero. So, the instantaneous velocity at time

t

v(t) = \lim_{h \to 0} v_{avg}(h) = \lim_{h \to 0} (5 - 2t - th + h^2)

. Evaluating this limit:

v(t) = 5 - 2t - t(0) + (0)^2 = 5 - 2t

. At time

t=4

v(4) = 5 - 2(4) = 5 - 8 = -3

. B is incorrect; this results from incorrectly treating the

th

term as

-4

instead of letting it go to zero in the limit. C is incorrect; this comes from taking the derivative of

v_{avg}

with respect to

h

and evaluating at

t=4

\frac{d}{dh}(-th + h^2) = -t + 2h = -4

when

t=4

and

h=0

. D is incorrect; this is the general formula for instantaneous velocity but not evaluated at the specific time

t=4

Question 17

For a function $f(x)$ , the expression $\frac{f(x) - f(3)}{x-3}$ represents the slope of a secant line through the points $(3, f(3))$ and $(x, f(x))$ . What does the quantity $\lim_{x \to 3} \frac{f(x) - f(3)}{x-3}$ represent?

The average rate of change of $f(x)$ over the interval $[x, 3]$ .
The change in the slope of the secant line as $x$ approaches $3$ .
The y-intercept of the tangent line to the graph of $f(x)$ at $x=3$ .
The slope of the tangent line to the graph of $f(x)$ at $x=3$ . (correct answer)

Explanation: When you see the expression

\frac{f(x) - f(3)}{x-3}

with a limit as

x

approaches 3, you're looking at the fundamental definition of a derivative. This is one of the most important limit forms in calculus. The expression

\frac{f(x) - f(3)}{x-3}

gives the slope of the secant line between points

(3, f(3))

and

(x, f(x))

. As

x

gets closer and closer to 3, this secant line approaches the tangent line at the point

(3, f(3))

. The limit

\lim_{x \to 3} \frac{f(x) - f(3)}{x-3}

captures the slope of that tangent line, which is precisely

f'(3)

- the derivative of

f

x = 3

. Choice A is incorrect because this limit doesn't represent an average rate of change over an interval. Instead, it's the instantaneous rate of change at a single point. Choice B misunderstands what's happening. The limit doesn't give us the change in slope of secant lines; it gives us what the slope approaches - the tangent line's slope. Choice C confuses slope with y-intercept. This limit tells us nothing about where the tangent line crosses the y-axis. Choice D correctly identifies that this limit represents the slope of the tangent line at

x = 3

. Study tip: Memorize this limit definition of the derivative:

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}

. When you see this pattern, you're always finding the slope of a tangent line, which equals the derivative at that point.

Question 18

Let $AROC(a, b)$ denote the average rate of change of a function $f(x)$ on the interval $[a, b]$ . If $f(x) = x^2 - x$ , what is $\lim_{h \to 0} AROC(2, 2+h)$ ?

0
2
3 (correct answer)
4

Explanation: The expression

\lim_{h \to 0} AROC(2, 2+h)

is the limit of the average rate of change over a shrinking interval around

x=2

. This is, by definition, the instantaneous rate of change (the derivative) of

f(x)

x=2

. First, find the derivative:

f'(x) = 2x - 1

. Then, evaluate it at

x=2

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

Question 19

Let $g(x) = \frac{1}{x+1}$ . Find the value $c$ in the interval $(0, 2)$ where the instantaneous rate of change of $g(x)$ at $x=c$ is equal to the average rate of change of $g(x)$ over the interval $[0, 2]$ .

$c = \sqrt{3} - 1$ (correct answer)
$c = \sqrt{3} + 1$
$c = \sqrt{3}$
$c = -1 - \sqrt{3}$

Explanation: First, calculate the average rate of change (AROC) of

g(x)

over

[0, 2]

\frac{g(2) - g(0)}{2 - 0} = \frac{1/3 - 1}{2} = \frac{-2/3}{2} = -\frac{1}{3}

. Next, find the instantaneous rate of change (IROC) by finding the derivative:

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

. Set the IROC at

c

equal to the AROC:

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

. This simplifies to

(c+1)^2 = 3

, so

c+1 = \pm\sqrt{3}

, which gives

c = -1 \pm \sqrt{3}

. Since

c

must be in the interval

(0, 2)

, we choose the positive root,

c = -1 + \sqrt{3} \approx 0.732

Question 20

Let $f(x)$ be a function such that $f''(x) > 0$ for all $x$ in the interval $[a, b]$ . Let $AROC$ be the average rate of change of $f$ over $[a,b]$ and let $IROC$ be the instantaneous rate of change of $f$ at the midpoint $c=\frac{a+b}{2}$ . Which statement is necessarily true?

$IROC > AROC$
$IROC < AROC$
$IROC \leq AROC$ (correct answer)
$IROC = AROC$

Explanation: The condition

f''(x) > 0

means the function

f(x)

is concave up. For a concave up function, the slope of the tangent line at the midpoint of an interval is always less than or equal to the slope of the secant line over that interval. The instantaneous rate of change (IROC) is the slope of the tangent line, and the average rate of change (AROC) is the slope of the secant line. Therefore,

IROC \leq AROC

. Equality holds for quadratic functions, so the strict inequality

IROC < AROC

is not always true.