Calculus 1 Quiz: Intro To Calculus Instantaneous Change

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Intro To Calculus Instantaneous ChangeQuestion 1 of 20

A particle's position is given by a differentiable function $p(t)$ . The average velocity of the particle over the time interval $[2, 5]$ is found to be 0. What can be definitively concluded from this information alone?

The particle was stationary for the entire interval

[2, 5]

The particle's instantaneous velocity is 0 at

t=2

and

t=5

The particle is at the same position at

t=2

and

t=5

The particle's maximum velocity on the interval

(2, 5)

is equal to its minimum velocity on that interval.

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Calculus 1 Quiz

Calculus 1 Quiz: Intro To Calculus Instantaneous Change

Practice Intro To Calculus Instantaneous Change 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 Intro To Calculus Instantaneous Change, 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 particle was stationary for the entire interval $[2, 5]$ .
The particle's instantaneous velocity is 0 at $t=2$ and $t=5$ .
The particle is at the same position at $t=2$ and $t=5$ . (correct answer)
The particle's maximum velocity on the interval $(2, 5)$ is equal to its minimum velocity on that interval.

Explanation: The correct answer is C. The average velocity over

[2, 5]

is given by

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

. If this value is 0, then the numerator must be 0, which means

p(5) - p(2) = 0

, or

p(5) = p(2)

. This means the particle's displacement is zero, so it is in the same position at the beginning and end of the interval. A is incorrect; the particle could have moved and returned to its starting position. B is incorrect; the particle could be moving at the endpoints. D implies the velocity is constant, which is not necessarily true; it only requires that the starting and ending positions are the same.

Question 2

The average rate of change of a function $f(x)$ on an interval $[a, b]$ corresponds to the slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$ . How is the instantaneous rate of change at $x = a$ conceptually derived from this?

By finding the average rate of change over the largest possible interval that contains $a$ .
By calculating the limit of the secant line slopes as the point $b$ approaches the point $a$ . (correct answer)
By determining the function's value $f(a)$ and dividing it by the point $a$ .
By finding the slope of a secant line that must also pass through the origin $(0,0)$ .

Explanation: The instantaneous rate of change is the limit of the average rate of change as the interval shrinks to a point. This is geometrically equivalent to finding the limit of the slopes of secant lines through

(a, f(a))

and

(b, f(b))

b

approaches

a

. The result is the slope of the tangent line at

x=a

Question 3

A particle's position along an axis is given by $p(t) = t^3 - 6t$ for $t \ge 0$ . Let $v_{avg,[a,b]}$ denote the average velocity over the interval $[a, b]$ . Which of the following correctly compares the average velocities over the intervals $[1, 2]$ and $[2, 3]$ ?

$v_{avg,[1,2]} > v_{avg,[2,3]}$
$v_{avg,[1,2]} < v_{avg,[2,3]}$ (correct answer)
$v_{avg,[1,2]} = v_{avg,[2,3]}$
The average velocity is negative over $[1, 2]$ and positive over $[2, 3]$ .

Explanation: First, calculate the average velocity for

[1, 2]

v_{avg,[1,2]} = \frac{p(2)-p(1)}{2-1}

. We have

p(2) = 2^3 - 6(2) = 8 - 12 = -4

and

p(1) = 1^3 - 6(1) = 1 - 6 = -5

. So,

v_{avg,[1,2]} = \frac{-4 - (-5)}{1} = 1

. Next, calculate the average velocity for

[2, 3]

v_{avg,[2,3]} = \frac{p(3)-p(2)}{3-2}

. We have

p(3) = 3^3 - 6(3) = 27 - 18 = 9

. So,

v_{avg,[2,3]} = \frac{9 - (-4)}{1} = 13

. Comparing the two values,

1 < 13

, so

v_{avg,[1,2]} < v_{avg,[2,3]}

Question 4

The temperature of a metal rod $x$ centimeters from a heated end is given by the function $T(x) = 300e^{-0.05x}$ degrees Celsius. Which of the following is the best interpretation of the value calculated by the expression $\frac{T(11) - T(10)}{11 - 10}$ ?

The exact rate of temperature change per centimeter at the point $x=10$ cm.
The average rate at which temperature changes with respect to position over the interval from $x=10$ cm to $x=11$ cm. (correct answer)
The total change in temperature, in degrees Celsius, between $x=10$ cm and $x=11$ cm.
The temperature at $x=10.5$ cm, which is the midpoint of the interval.

Explanation: The expression

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

represents the average rate of change of the function

f

over the interval

[a,b]

. In this context, it calculates the average change in temperature per centimeter over the interval from

x=10

x=11

. Choice A describes the instantaneous rate of change,

T'(10)

. Choice C describes the numerator only,

T(11)-T(10)

. Choice D describes the function value at the midpoint,

T(10.5)

Question 5

Which of the following limits represents the instantaneous rate of change of a function $f(x)$ at a point $x=a$ ?

$\lim_{h \to 0} \frac{\cos(\pi + h) - \cos(\pi)}{h}$ (correct answer)
$\lim_{h \to 0} \frac{\tan(h)}{h}$
$\lim_{x \to \pi} \frac{\sin(x) - \sin(\pi)}{x}$
$\lim_{h \to \pi} \frac{e^{x+h} - e^{x}}{h}$

Explanation: The definition of the instantaneous rate of change (the derivative) of a function

f

at a point

a

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

. Option A matches this definition for the function

f(x) = \cos(x)

at the point

a = \pi

. Option B represents the derivative of

f(x)=\tan(x)

x=0

, as

\tan(0)=0

. Option C is an incorrect form of the alternative definition; it should divide by

x-\pi

. Option D is incorrect because the limit must be taken as

h \to 0

, not

h \to \pi

Question 6

Consider the function $f(x) = x^2$ , for which the instantaneous rate of change at $x=1$ is 2. Let $m_{sec}$ be the slope of the secant line connecting the points $(1, f(1))$ and $(b, f(b))$ for $b \neq 1$ . Which statement correctly describes $m_{sec}$ ?

For $b > 1$ , $m_{sec} < 2$ , and for $b < 1$ , $m_{sec} > 2$ .
For $b > 1$ , $m_{sec} > 2$ , and for $b < 1$ , $m_{sec} < 2$ . (correct answer)
$m_{sec} = 2$ for all values of $b \neq 1$ .
$m_{sec} < 2$ for all values of $b \neq 1$ .

Explanation: The slope of the secant line is

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

. The instantaneous rate of change at

x=1

is 2. If

b > 1

, then

m_{sec} = b+1 > 1+1 = 2

. If

b < 1

, then

m_{sec} = b+1 < 1+1 = 2

. This is consistent with the graph of the concave up parabola

y=x^2

, where secant lines from the right of a point are steeper than the tangent, and secant lines from the left are less steep.

Question 7

The rate of change of a function $f(x)$ at $x=c$ is given by the limit $L = \lim_{h \to 0} \frac{(c+h)^3 - c^3}{h}$ . If $c > 0$ , what is the geometric interpretation of the value $\frac{(c+0.1)^3 - c^3}{0.1}$ ?

The exact slope of the tangent line to the graph of $y=x^3$ at $x=c$ .
The height of the function $y=x^3$ at the point $x=c+0.1$ .
The slope of the secant line to the graph of $y=x^3$ between $x=c$ and $x=c+0.1$ . (correct answer)
The area under the curve $y=x^3$ from $x=c$ to $x=c+0.1$ .

Explanation: When you encounter expressions involving

\frac{f(x+h) - f(x)}{h}

, you're looking at the foundation of derivatives - the difference quotient. This formula measures how much a function changes over a small interval, which connects directly to geometric interpretations on graphs. The expression

\frac{(c+0.1)^3 - c^3}{0.1}

follows this pattern exactly. Here, you're taking two points on the curve

y = x^3

: the point

(c, c^3)

and the point

(c+0.1, (c+0.1)^3)

. The numerator gives you the vertical change (rise) between these points, while the denominator gives you the horizontal change (run). This rise-over-run calculation is precisely the slope of the secant line connecting these two points, making C correct. Option A is incorrect because the exact slope of the tangent line requires taking the limit as

h \to 0

. With

h = 0.1

, you're approximating the tangent slope, not calculating it exactly. Option B is wrong because

(c+0.1)^3

by itself would give you the height at

x = c+0.1

, but the entire difference quotient represents a rate of change, not a function value. Option D is incorrect because areas under curves involve integration (antiderivatives), not difference quotients. Remember this key distinction: when

h

has a specific non-zero value, you're finding a secant line slope. Only when

h \to 0

do you get the tangent line slope. This difference between approximation and exact calculation appears frequently in calculus problems.

Question 8

Let $v(t)$ be the velocity of a particle at time $t$ . The expression $\frac{v(5)-v(2)}{3}$ represents the particle's average acceleration from $t=2$ to $t=5$ . If $v(t)$ is a non-linear function, which statement best describes the relationship between this value and the instantaneous acceleration at $t=2$ , denoted $a(2)$ ?

The average acceleration approximates $a(2)$ , but differs because the interval $[2, 5]$ is not infinitesimal. (correct answer)
The average acceleration is always equal to the instantaneous acceleration $a(2)$ for any function.
The average acceleration equals the average of instantaneous accelerations at $t=2$ and $t=5$ .
The average acceleration has different units than instantaneous acceleration, preventing comparison.

Explanation: When you encounter questions about average versus instantaneous rates of change, you're dealing with a fundamental calculus concept: the relationship between secant and tangent lines. The expression

\frac{v(5)-v(2)}{3}

calculates the average acceleration over the interval [2,5] using the slope of the secant line connecting points (2,v(2)) and (5,v(5)). Meanwhile, instantaneous acceleration

a(2) = v'(2)

represents the slope of the tangent line at exactly

t=2

. Answer A is correct because average acceleration provides an approximation of instantaneous acceleration, but they differ when the interval isn't infinitesimally small. For non-linear velocity functions, the secant line slope varies depending on the interval length. As the interval shrinks toward zero, the average acceleration approaches the instantaneous value—this is precisely the definition of a derivative. Answer B is wrong because average and instantaneous accelerations are only equal for linear velocity functions (constant acceleration). With non-linear functions, they differ unless the interval approaches zero. Answer C incorrectly suggests the average acceleration equals

\frac{a(2)+a(5)}{2}

. This confuses the average rate of change with the average of rates at endpoints—these are entirely different calculations. Answer D is incorrect because both average and instantaneous acceleration have identical units (velocity per time, like m/s²), making comparison perfectly valid. Remember: average rates use secant lines over intervals, while instantaneous rates use tangent lines at points. They're related through the limit process that defines derivatives.

Question 9

The instantaneous rate of change of the function $f(x)=1/x$ at $x=2$ is $-1/4$ . Which of the following intervals $[2, b]$ would produce an average rate of change closest to $-1/4$ ?

$[2, 4]$
$[2, 3]$
$[2, 2.1]$
$[2, 2.01]$ (correct answer)

Explanation: The instantaneous rate of change is the limit of the average rate of change as the interval length approaches zero. To get an average rate of change that is closest to the instantaneous rate, one must choose the smallest possible interval. The lengths of the intervals are

4-2=2

3-2=1

2.1-2=0.1

, and

2.01-2=0.01

. The interval

[2, 2.01]

is the smallest, so its average rate of change will be the best approximation of the instantaneous rate of change.

Question 10

The expression $\lim_{h \to 0} \frac{s(t+h) - s(t)}{h}$ describes the instantaneous velocity of an object at time $t$ , where $s(t)$ is the position function. Which statement best explains why this limit is necessary to define instantaneous velocity?

The formula for average velocity $\frac{\Delta s}{\Delta t}$ is undefined when the time interval $\Delta t$ is zero, so the limit provides a way to compute a value by simply setting $\Delta t$ to a very small number.
Instantaneous velocity is the change in position over the smallest possible non-zero time interval, and the limit is the mathematical tool used to find the value of this interval.
It represents the unique value that the average velocities over shrinking time intervals $[t, t+h]$ approach, capturing the trend of the rate of change at the single moment $t$ . (correct answer)
The limit process calculates the arithmetic mean of all possible average velocities for intervals starting at $t$ to find the most probable rate of change at that instant.

Explanation: The correct answer is C. Instantaneous velocity cannot be calculated directly because at a single instant,

\Delta t = 0

, leading to division by zero. Instead, we examine the behavior of the average velocity over progressively smaller time intervals starting at

t

. The limit, if it exists, is the value these average velocities converge to, representing the rate of change at that precise moment. A is incorrect because the limit is not about plugging in a small number, but about a trend as the interval size approaches zero. B is incorrect because there is no 'smallest possible non-zero' time interval in the real number system. D is incorrect as a limit is a process of convergence to a unique value, not an averaging process.

Question 11

A student incorrectly attempts to find the instantaneous rate of change of $f(x) = x^2$ at $x=3$ by directly substituting $h=0$ into the expression $\frac{f(3+h) - f(3)}{h}$ . What is the mathematical reason this approach fails?

The approach fails because it calculates the average rate of change, not the instantaneous rate of change.
The approach fails because it leads to the indeterminate form $\frac{0}{0}$ , which provides no information about the rate of change. (correct answer)
The approach fails because $h$ must be a small positive number, not exactly zero, for the formula to be valid.
The approach fails because $f(3+0)$ is not a valid mathematical operation for finding a rate of change.

Explanation: The correct answer is B. Substituting

h=0

gives

\frac{f(3) - f(3)}{0} = \frac{0}{0}

. This is an indeterminate form, meaning the expression is undefined and the result cannot be determined from the form alone. The entire point of using a limit is to determine what value the expression approaches as

h

gets arbitrarily close to 0, precisely because direct substitution is not possible. A is incorrect; the expression with a non-zero

h

is the average rate of change, but substituting

h=0

does not calculate this. C is a common misconception; the limit considers

h

approaching 0 from both positive and negative sides. D is incorrect;

f(3+0)

is just

f(3)

, which is a valid operation, but this does not address the division by zero.

Question 12

The temperature $T$ (in degrees Celsius) of a chemical reaction is measured at different times $t$ (in seconds). The following data are collected: $T(1.0) = 40.0$ , $T(1.1) = 42.0$ , $T(1.2) = 43.5$ , and $T(2.0) = 55.0$ .

Based on the available data, what is the best estimate for the instantaneous rate of change of temperature at $t=1.1$ seconds?

15.0 °C/s
16.7 °C/s
17.5 °C/s (correct answer)
20.0 °C/s

Explanation: The correct answer is C. To estimate the instantaneous rate of change at a point, we should calculate the average rate of change over the smallest available interval that contains that point. Here, the smallest interval around

t=1.1

for which we have data is

[1.0, 1.2]

. The average rate of change over this interval is

\frac{T(1.2) - T(1.0)}{1.2 - 1.0} = \frac{43.5 - 40.0}{0.2} = \frac{3.5}{0.2} = 17.5

°C/s. Answer D is the average rate over

[1.0, 1.1]

(

\frac{42-40}{0.1}=20

), and Answer A is the average rate over

[1.1, 1.2]

(

\frac{43.5-42}{0.1}=15

). While both are reasonable, the average over the symmetric interval

[1.0, 1.2]

is generally a better estimate. Answer B is the average rate over the large interval

[1.2, 2.0]

, which is not a good estimate for the rate at

t=1.1

. Using the symmetric interval is a standard technique for numerical estimation.

Question 13

The number of bacteria in a culture is given by the function $P(t)$ , where $t$ is measured in hours. The value of the expression $\frac{P(4) - P(2)}{2}$ was calculated to be 1000. What is the most accurate interpretation of this result?

The instantaneous growth rate of the bacteria population at $t=3$ was exactly 1000 bacteria per hour.
Between the second and fourth hour, the bacteria population increased at an average rate of 1000 bacteria per hour. (correct answer)
The bacteria population at $t=3$ was exactly 1000.
The rate of growth of the bacteria population at $t=2$ was exactly 1000 bacteria per hour.

Explanation: The correct answer is B. The expression

\frac{P(4) - P(2)}{4 - 2}

is the definition of the average rate of change of the function

P(t)

over the interval

[2, 4]

. Therefore, 1000 represents the average increase in bacteria per hour during this two-hour period. A is incorrect because this is an average rate, not an instantaneous rate. While the instantaneous rate at

t=3

might be close to 1000, it is not guaranteed to be exactly 1000. C is incorrect; the value represents a rate of change, not the population size. D is incorrect as it refers to the instantaneous rate at

t=2

, not the average rate over an interval.

Question 14

The concept of instantaneous rate of change describes how a quantity is changing at a single moment in time. This presents an apparent paradox: at a single instant, no time passes, so how can change occur?

Which statement best resolves this apparent paradox from the perspective of standard calculus?

Instantaneous change is a theoretical ideal; in physical reality, change only occurs over measurable, non-zero intervals of time.
The paradox is resolved by defining instantaneous rate not as a change over zero time, but as the limit of average rates of change as the time interval's duration shrinks towards zero. (correct answer)
At an instant, the function possesses a 'tendency to change' which is measured by the derivative, even if no change actually happens in zero time.
The paradox is resolved by using infinitesimally small time steps, called $dt$ , which are not zero but are smaller than any real number, allowing for a non-zero change.

Explanation: The correct answer is B. Standard calculus, based on the concept of limits, resolves the paradox by not attempting to calculate change over a zero-duration interval. Instead, it defines the instantaneous rate as the value that average rates (which are well-defined) approach as the interval duration becomes arbitrarily close to zero. A is a philosophical argument but not the mathematical resolution. C uses vague, non-mathematical language ('tendency to change') that doesn't align with the formal definition. D describes the framework of non-standard analysis, which uses infinitesimals; while a valid approach to calculus, it is not the standard limit-based approach taught in a typical Calculus I course.

Question 15

Suppose for a function $g(t)$ , we know that the average rate of change over the interval $[3, 3+h]$ is given by the expression $5h^2 - 2h + 7$ . What is the instantaneous rate of change of $g(t)$ at $t=3$ ?

7 (correct answer)
5
0
The instantaneous rate of change cannot be determined from the given information.

Explanation: This question tests your understanding of the relationship between average and instantaneous rates of change, which connects to the fundamental definition of the derivative. When you know the average rate of change over an interval

[a, a+h]

, you can find the instantaneous rate of change at

t = a

by taking the limit as

h

approaches 0. The average rate of change is given by

\frac{g(3+h) - g(3)}{h} = 5h^2 - 2h + 7

. To find the instantaneous rate of change at

t = 3

, you need:

\lim_{h \to 0} (5h^2 - 2h + 7)

Substituting

h = 0

5(0)^2 - 2(0) + 7 = 7

. Looking at the wrong answers: Answer B (5) might tempt you if you mistakenly focused on the coefficient of the

h^2

term, but this coefficient doesn't determine the limit. Answer C (0) could result from incorrectly thinking that limits always equal zero or from confusing this with finding where the derivative equals zero. Answer D suggests the information is insufficient, but having the average rate of change formula actually gives you exactly what you need to find the instantaneous rate. The correct answer is A) 7. Study tip: Remember that instantaneous rate of change is the limit of average rates of change as the interval shrinks to zero. When you have an algebraic expression for the average rate of change over

[a, a+h]

, simply substitute

h = 0

to find the instantaneous rate at

t = a

Question 16

Which of the following expressions, where $h$ is a small non-zero number, would typically provide the most accurate estimate of the instantaneous rate of change of a differentiable function $f$ at $x=a$ ?

$\frac{f(a+h) - f(a-h)}{2h}$ (a central difference) (correct answer)
$\frac{f(a) - f(a-h)}{h}$ (a backward difference)
$\frac{f(a+h) - f(a)}{h}$ (a forward difference)
$\frac{f(a+h) + f(a-h)}{2}$ (an average value)

Explanation: When you encounter questions about estimating derivatives numerically, you're dealing with different approximation methods for the instantaneous rate of change. The key insight is understanding how these approximations relate to the formal definition of a derivative:

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

The central difference formula in choice A,

\frac{f(a+h) - f(a-h)}{2h}

, provides the most accurate estimate because it's geometrically superior to one-sided approaches. This formula calculates the slope of the secant line connecting points at equal distances on both sides of

x = a

. As

h

approaches zero, this secant line approaches the tangent line more accurately than one-sided methods. Choice B uses a backward difference, estimating the derivative using only points to the left of

a

. While this converges to the correct derivative as

h \to 0

, it's less accurate for finite values of

h

because it only uses information from one side. Choice C employs a forward difference, using only points to the right of

a

. Like the backward difference, this is one-sided and therefore less accurate than the central approach for practical calculations. Choice D calculates

\frac{f(a+h) + f(a-h)}{2}

, which gives an average of function values, not a rate of change. This expression has no

h

in the denominator and doesn't approximate a derivative at all. Remember: central difference formulas typically provide better numerical approximations because they use symmetric information around the point of interest, reducing approximation error compared to one-sided methods.

Question 17

The average velocity of a projectile on the interval $[1, 4]$ is 20 m/s. The average velocity on the interval $[4, 7]$ is 40 m/s. What is the average velocity of the projectile over the entire interval $[1, 7]$ ?

25 m/s
The average velocity cannot be determined without the position function.
35 m/s
30 m/s (correct answer)

Explanation: When you encounter average velocity problems over multiple intervals, remember that average velocity equals total displacement divided by total time. The key insight is that you can find the total displacement by adding up the displacements from each subinterval. For the interval [1, 4], the time elapsed is 3 seconds and average velocity is 20 m/s, so the displacement is

20 \times 3 = 60

meters. For the interval [4, 7], the time elapsed is also 3 seconds and average velocity is 40 m/s, giving a displacement of

40 \times 3 = 120

meters. The total displacement over [1, 7] is

60 + 120 = 180

meters, and the total time is

6

seconds. Therefore, the average velocity is

\frac{180}{6} = 30

m/s, which is answer D. Let's examine why the other options are incorrect. Choice A (25 m/s) represents the simple arithmetic mean of 20 and 40, but this ignores that average velocity depends on total displacement and total time, not just averaging the given velocities. Choice C (35 m/s) might come from incorrectly weighting the velocities or making an arithmetic error. Choice B is wrong because you don't need the actual position function—average velocity problems can be solved using just the definition of average velocity and the given information about each interval. Remember: for average velocity over combined intervals, always calculate total displacement and divide by total time. Don't just average the given velocities unless the time intervals are your primary concern.

Question 18

The average rate of change of a function $f(x)$ over the interval $[x, x+h]$ is $\frac{\Delta f}{\Delta x}$ . The instantaneous rate of change is $\lim_{h \to 0} \frac{\Delta f}{\Delta x}$ . If $f(x)$ is a linear function, say $f(x) = mx+b$ , what is the relationship between these two rates?

The average rate of change is equal to the instantaneous rate of change for any choice of $x$ and $h$ . (correct answer)
The average rate of change approaches the instantaneous rate of change only when $h=0$ .
The instantaneous rate of change is always greater than the average rate of change.
The instantaneous rate of change is the average of all possible average rates of change.

Explanation: This question tests your understanding of rates of change and what makes linear functions special. When you encounter problems comparing average and instantaneous rates of change, always consider the function's shape—this determines whether these rates differ. For a linear function

f(x) = mx + b

, let's calculate both rates. The average rate of change over

[x, x+h]

is:

\frac{\Delta f}{\Delta x} = \frac{f(x+h) - f(x)}{h} = \frac{[m(x+h) + b] - [mx + b]}{h} = \frac{mh}{h} = m

The instantaneous rate of change is:

\lim_{h \to 0} \frac{\Delta f}{\Delta x} = \lim_{h \to 0} m = m

Both equal

m

, the slope of the line, regardless of the values of

x

and

h

. Answer A is correct because linear functions have constant slope everywhere—the average rate over any interval equals the instantaneous rate at any point. Answer B is wrong because the rates are equal for all values of

h

, not just when

h = 0

(which would make

\frac{\Delta f}{\Delta x}

undefined anyway). Answer C is incorrect since we've shown both rates equal

m

—neither is greater than the other. Answer D misrepresents the relationship entirely. The instantaneous rate isn't an average of other rates; it's the limit of average rates as the interval shrinks. Study tip: Linear functions are unique because their rate of change is constant. For any curved function, average and instantaneous rates will differ, but straight lines have the same "steepness" everywhere.

Question 19

Let $f(x) = x^2$ . The average rate of change over the interval $[1, 1+h]$ is $2+h$ . The instantaneous rate of change at $x=1$ is 2. For any $h>0$ , the average rate of change is an overestimate of the instantaneous rate. What feature of the graph of $f(x)=x^2$ on $[1, \infty)$ explains this?

The function is concave up on this interval. (correct answer)
The function is always increasing on this interval.
The function is always positive on this interval.
The function grows without bound on this interval.

Explanation: This question tests your understanding of how concavity affects the relationship between average and instantaneous rates of change. When you see problems comparing these rates, think about the shape of the curve between the two points. The key insight is that concavity determines whether secant lines (which give average rates of change) lie above or below the curve. For

f(x) = x^2

[1, \infty)

, the function is concave up, meaning it curves upward like a smile. When a function is concave up, any secant line connecting two points on the curve lies above the curve between those points. Since the secant line from

(1, f(1))

(1+h, f(1+h))

lies above the curve, its slope (the average rate of change

2+h

) is greater than the slope of the tangent line at

x=1

(the instantaneous rate of change, which is 2). This explains why the average rate consistently overestimates the instantaneous rate. Looking at the wrong answers: B is true but irrelevant—a function can be increasing while its average rate underestimates the instantaneous rate if it's concave down. C is also true but doesn't explain the overestimate relationship; the sign of function values has no bearing on rate comparisons. D is true since

x^2 \to \infty

x \to \infty

, but unbounded growth alone doesn't determine whether average rates overestimate or underestimate instantaneous rates. Remember: concave up means secant lines lie above the curve, making average rates of change overestimate instantaneous rates. Concave down gives the opposite effect.

Question 20

Consider the function $f(x) = \begin{cases} x^2 & \text{if } x \le 1 \\ 3x-2 & \text{if } x > 1 \end{cases}$ . A student computes the limit of the difference quotient from the left of $x=1$ and from the right of $x=1$ . What will they find about the instantaneous rate of change at $x=1$ ?

It exists and is 2, because the rate of change of $x^2$ at $x=1$ is 2.
It exists and is 3, because the rate of change of $3x-2$ for $x>1$ is 3.
It does not exist, because the function is not continuous at $x=1$ .
It does not exist, because the rate of change from the left is 2 and from the right is 3. (correct answer)

Explanation: The correct answer is D. First, check for continuity at

x=1

f(1)=1^2=1

\lim_{x\to 1^+} (3x-2) = 3(1)-2=1

. The function is continuous. Now, we check the limits of the difference quotient from both sides. From the left (using

x^2

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

. From the right (using

3x-2

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

. Since the left-hand derivative (2) does not equal the right-hand derivative (3), the instantaneous rate of change (the derivative) does not exist at

x=1

. This signifies a 'corner' in the graph.