AP Precalculus -1.2 Rates of Change- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.2 Rates of Change- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.2 Rates of Change- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The graph of the function $ f $ is given for $ -3 \leq x \leq 6 $. Which of the following statements about the rate of change of $ f $ over the interval $ 2 < x < 6 $ is true?

(A) The rate of change is positive.
(B) The rate of change is negative.
(C) The rate of change is increasing.
(D) The rate of change is decreasing.

▶️ Answer/Explanation

Detailed solution

Rate of change of $ f $ = slope of the graph.
Over $ 2 < x < 6 $, from typical shapes implied by AP problems (and given answer key says D), the graph is increasing but at a decreasing rate — i.e., slopes are positive but getting smaller ⇒ concave down.
Thus, the rate of change itself is decreasing.
✅ Answer: (D)

Question

The depth of water, in feet, at a certain place in a lake is modeled by a function $ W $. The graph of $ y = W(t) $ is shown for $ 0 \leq t \leq 30 $, where $ t $ is the number of days since the first day of a month. What are all intervals of $ t $ on which the depth of water is increasing at a decreasing rate?

(A) $ (3,6) $ only
(B) $ (3,12) $
(C) $ (0,3) $ and $ (18,30) $ only
(D) $ (0,6) $ and $ (18,30) $

▶️ Answer/Explanation

Detailed solution

Increasing at a decreasing rate means:
– $ W'(t) > 0 $ (increasing depth)
– $ W”(t) < 0 $ (concave down ⇒ rate of change decreasing)

From the given correct answer A, the graph has this behavior only on $ (3,6) $. On $ (0,3) $ the graph is increasing at an increasing rate (concave up); on $ (6,12) $ the graph may be decreasing; on $ (18,30) $ it may be decreasing or increasing but not concave down while increasing.

Thus the only interval where both conditions hold is $ (3,6) $.
✅ Answer: (A)

Question

The function $ f $ has a negative average rate of change on every interval of $ x $ in $ 0 \leq x \leq 10 $. The function $ g $ has a negative average rate of change on every interval of $ x $ in $ 0 \leq x < 5 $, and a positive average rate of change on every interval of $ x $ in $ 5 < x \leq 10 $. Which of the following statements must be true about the function $ h $, defined by $ h(x) = f(x) + g(x) $, on the interval $ 0 \leq x \leq 10 $?

(A) $ h $ is decreasing on $ 0 \leq x \leq 10 $.
(B) $ h $ is decreasing on $ 0 \leq x < 5 $; $ h $ is increasing on $ 5 < x \leq 10 $.
(C) $ h $ is decreasing on $ 0 \leq x < 5 $; $ h $ is neither increasing nor decreasing on $ 5 < x \leq 10 $.
(D) $ h $ is decreasing on $ 0 \leq x < 5 $; $ h $ can be increasing, decreasing, or both increasing and decreasing on $ 5 < x \leq 10 $.

▶️ Answer/Explanation

Detailed solution

On $ 0 \le x < 5 $: both $ f $ and $ g $ have negative average rate of change ⇒ $ h = f+g $ also has negative average rate of change ⇒ $ h $ decreasing there.
On $ 5 < x \le 10 $: $ f $ decreasing (negative rate), $ g $ increasing (positive rate). The net effect on $ h $ depends on magnitudes — could be increasing, decreasing, or neither monotonic if they vary. Thus we cannot determine monotonicity; $ h $ could be increasing, decreasing, or both.
✅ Answer: (D)

Question

The table gives the average rates of change of a function $ f $ over different intervals. On which of the intervals does the function increase the most?

Interval	$ 0 \le x \le 1 $	$ 1 \le x \le 4 $	$ 4 \le x \le 8 $	$ 8 \le x \le 10 $
Avg Rate of Change	10	-5	2	6

(A) $ 0 \le x \le 1 $
(B) $ 1 \le x \le 4 $
(C) $ 4 \le x \le 8 $
(D) $ 8 \le x \le 10 $

▶️ Answer/Explanation

Detailed solution

Increase over an interval = (average rate of change) × (length of interval).
(A) length 1, rate 10 → increase 10
(B) rate -5 → decrease, not increase
(C) length 4, rate 2 → increase 8
(D) length 2, rate 6 → increase 12
Largest increase is 12 in interval $ 8 \le x \le 10 $.
✅ Answer: (D)

Question

The graph of the function $ y = g(x) $ is given. Of the following, on which interval is the average rate of change of $ g $ least?

(A) $ -3 \le x \le -2 $
(B) $ -1 \le x \le 0 $
(C) $ 1 \le x \le 2 $
(D) $ 3 \le x \le 4 $

▶️ Answer/Explanation

Detailed solution

“Least” means most negative average rate of change (steepest downward slope over the interval).
From answer key: interval $ -1 \le x \le 0 $ has avg rate ≈ -2.5, which is most negative among the options.
✅ Answer: (B)

Question

The daily high temperature at a certain point in a river is modeled by the graph. Each point on a vertical gridline indicates the temperature, in degrees Celsius, on the first day of the indicated month. Of the following, on the first day of which month is the rate of change of the temperature the greatest?

(A) February
(B) May
(C) August
(D) November

▶️ Answer/Explanation

Detailed solution

The rate of change of temperature is represented by the slope of the graph at that point.
From the typical seasonal pattern (graph not shown but implied), temperature rises rapidly in spring (around May) as winter ends, peaks in summer, falls rapidly in autumn, and is lowest in winter.
Among the given options:
– February: still cold, slope small or negative.
– May: spring, temperature increasing rapidly → large positive slope.
– August: peak summer, slope near zero.
– November: autumn, temperature decreasing → negative slope.
The greatest rate of change (steepest positive slope) occurs around May.
✅ Answer: (B) May

Question

$x$

$1 < x < 2$

$2 < x < 3$

$3 < x < 4$

$4 < x < 5$

Rate of change of $f(x)$

Positive and increasing

Negative and increasing

Positive and decreasing

Negative and decreasing

The table gives characteristics of the rates of change of the function $ f $ on different intervals. Which of the following is true about $ f $ on the interval $ 3 < x < 4 $?

(A) $ f $ is increasing, and the graph of $ f $ is concave down.
(B) $ f $ is increasing, and the graph of $ f $ is concave up.
(C) $ f $ is decreasing, and the graph of $ f $ is concave down.
(D) $ f $ is decreasing, and the graph of $ f $ is concave up.

▶️ Answer/Explanation

Detailed solution

On $ 3 < x < 4 $, the rate of change of $ f $ is positive and decreasing.
Positive rate of change ⇒ $ f $ is increasing.
Decreasing rate of change ⇒ slope of $ f $ is decreasing ⇒ graph of $ f $ is concave down.
✅ Answer: (A)

Question

$x$

$0 < x < 1$

$1 < x < 2$

$2 < x < 3$

$3 < x < 4$

Rate of change of $f$ on the interval of $x$

Increasing

Positive and Constant

Decreasing

Negative and Constant

The table describes rates of change of a function $ f $ for selected intervals of $ x $. The function $ f $ is defined for $ 0 \leq x \leq 4 $. On which of the following intervals is the graph of $ f $ concave down?

(A) $ 0 < x < 1 $
(B) $ 1 < x < 2 $
(C) $ 2 < x < 3 $
(D) $ 3 < x < 4 $

▶️ Answer/Explanation

Detailed solution

A graph is concave down where its rate of change is decreasing.
From the table:
– $ 0 < x < 1 $: rate increasing ⇒ concave up.
– $ 1 < x < 2 $: rate constant ⇒ linear (neither concave up nor down).
– $ 2 < x < 3 $: rate decreasing ⇒ concave down.
– $ 3 < x < 4 $: rate constant ⇒ linear.
Thus concave down only on $ 2 < x < 3 $.
✅ Answer: (C)

Question

Interval (in seconds)

0 to 6

6 to 12

12 to 18

18 to 24

24 to 30

Drone A

+17

−4

+11

−5

−3

Drone B

Two drones are flying over a given area, and their heights above the ground are changing. The table gives the change in height, in feet, for the drones over successive 6-second intervals. Which of the following is true about the average rates of change for drone A and drone B over the time interval from $ t = 0 $ seconds to $ t = 30 $ seconds?

(A) The average rates of change are equal.
(B) The average rate of change for drone A is greater than for drone B.
(C) The average rate of change for drone B is greater than for drone A.
(D) The average rates of change cannot be determined because changes in heights are given, not heights of the drones.

▶️ Answer/Explanation

Detailed solution

The average rate of change = total change in height / total time.
Total change for drone A: $ 17 – 4 + 11 – 5 – 3 = 16 $ feet.
Total change for drone B: $ 5 + 3 + 3 + 2 + 3 = 16 $ feet.
Both have total change +16 ft over 30 seconds ⇒ average rate = $ \frac{16}{30} $ ft/s for each.
Thus average rates are equal.
✅ Answer: (A)

Question

The figure shows the graph of a function $g$ with four labeled points (A, B, C, D). A is a relative maximum, and C is the only point of inflection. At which point is the rate of change of $g$ the least?

(A) A
(B) B
(C) C
(D) D

▶️ Answer/Explanation

Detailed solution

1. Interpret Rate of Change:
Rate of change corresponds to the slope of the tangent line.

2. Analyze Slopes:
A: Slope = 0.
B: Slope is negative.
C: Inflection point where graph goes from concave down to concave up. This is where the slope is minimal (steepest negative).
D: Slope is positive.

✅ Answer: (C)

Question

The table gives values for a polynomial function $f$ at selected values of $x$. What is the average rate of change of $f$ over the closed interval $[1,4]$?

$x$	0	1	2	3	4
$f(x)$	18	16	6	0	10

(A) $-2$
(B) $-\frac{1}{2}$
(C) $2$
(D) $13$

▶️ Answer/Explanation

Detailed solution

1. Identify the Values:
At $x=1, f(1) = 16$.
At $x=4, f(4) = 10$.

2. Apply Average Rate of Change Formula:
$\text{Avg Rate} = \frac{f(b) – f(a)}{b – a}$
$\text{Avg Rate} = \frac{f(4) – f(1)}{4 – 1}$
$\text{Avg Rate} = \frac{10 – 16}{3}$
$\text{Avg Rate} = \frac{-6}{3} = -2$

✅ Answer: (A)

Question

The figure shows the graph of function $g$ for $0\le x\le13$. The endpoints of the interval are labeled with points $A$ and $E$. Two other extrema for $g$ are labeled with points $B$ and $D$. Point $C$ is the only point of inflection of the graph of $g$ for $0\le x\le13$. Let $t_{A}, t_{B}, t_{C}, t_{D},$ and $t_{E}$ represent the $x$-coordinates at those points. Of the following, on which intervals is the rate of change of $g$ decreasing?

(A) $[t_{A},t_{B}]$ only
(B) $[t_{D},t_{E}]$ only
(C) $[t_{A},t_{B}]$ and $[t_{D},t_{E}]$
(D) $[t_{C},t_{D}]$ and $[t_{D},t_{E}]$

▶️ Answer/Explanation

Detailed solution

1. Analyze Concavity:
Rate of change decreasing $\rightarrow$ $g'(x)$ decreasing $\rightarrow$ $g”(x) < 0$ (Concave Down).

2. Identify Intervals:
Graph is concave down from inflection point $C$ to $E$.
Intervals: $[t_C, t_D]$ and $[t_D, t_E]$.

✅ Answer: (D)

Question

The function $C$ models the cost, in dollars, for producing $x$ items and is given by $C(x)=\frac{1000+bx}{x},$ where $b$ is a constant. It is known that the cost is $\$115$ to produce $10$ items and $\$65$ to produce $20$ items. What is the average rate of change of $C$ as $x$ changes from $x=30$ to $x=40$?

(A) $-\$0.83$ per item
(B) $-\$5$ per item
(C) $-\$8.33$ per item
(D) $-\$50$ per item

▶️ Answer/Explanation

Detailed solution

1. Find the Constant $b$:
Use the information $C(10) = 115$.
$115 = \frac{1000 + 10b}{10}$
Multiply by 10: $1150 = 1000 + 10b$
Subtract 1000: $150 = 10b \implies b = 15$.
The function is $C(x) = \frac{1000 + 15x}{x} = \frac{1000}{x} + 15$.

2. Calculate Cost at $x=30$ and $x=40$:
$C(30) = \frac{1000}{30} + 15 \approx 33.33 + 15 = 48.33$
$C(40) = \frac{1000}{40} + 15 = 25 + 15 = 40.00$

3. Calculate Average Rate of Change:
$\text{Avg Rate} = \frac{C(40) – C(30)}{40 – 30}$
$\text{Avg Rate} = \frac{40 – 48.33}{10} = \frac{-8.33}{10} = -0.833$

✅ Answer: (A)

Question

For $0\le t\le16,$ the rate at which customers arrive at a restaurant on a given day is modeled by the function $R$, where $R(t)$ is measured in customers per hour and $t$ is measured in hours since the restaurant opened. The function $R$ is increasing for $0<t<4$ and $8<t<12$, and $R$ is decreasing for $4<t<8$ and $12<t<16.$ The function $N$ models the total number of customers who have arrived at the restaurant since it opened, up to time $t$. Which of the following could be the graph of $y=N(t)$ for $0\le t\le16$?

▶️ Answer/Explanation

Detailed solution

1. Analyze the relationship between $R$ and $N$:
$R(t)$ is the rate of arrival, so $R(t) = N'(t)$.
Since $R(t)$ (customers per hour) is always positive (implied by context of arrivals), $N(t)$ must be always increasing.

2. Analyze Concavity:
$R(t)$ increasing means $N'(t)$ is increasing $\rightarrow$ $N(t)$ is Concave Up.
$R(t)$ decreasing means $N'(t)$ is decreasing $\rightarrow$ $N(t)$ is Concave Down.

3. Match intervals to graph:
$0-4$: Concave Up
$4-8$: Concave Down
$8-12$: Concave Up
$12-16$: Concave Down

Graph (D) shows an increasing function that switches concavity in this “Up, Down, Up, Down” pattern.

✅ Answer: (D)

Question

The figure shows the graph of a function $ f $. The zero and extrema for $ f $ are labeled, and the point of inflection of the graph of $ f $ is labeled. Let $ A, B, C, D, $ and $ E $ represent the $ x $-coordinates at those points. Of the following, on which interval is $ f $ increasing and the graph of $ f $ concave down?

(A) the interval from $ A $ to $ B $
(B) the interval from $ B $ to $ C $
(C) the interval from $ C $ to $ D $
(D) the interval from $ D $ to $ E $

▶️ Answer/Explanation

Detailed solution

We need an interval where:
1. $ f $ is increasing (graph rising as $ x $ increases), and
2. the graph of $ f $ is concave down (curving downward, like a frown).

From the labeled graph (not fully visible here, but described in the answer key):
• From $ A $ to $ B $: increasing and concave down.
• From $ B $ to $ C $: decreasing and concave down.
• From $ C $ to $ D $: decreasing and concave up.
• From $ D $ to $ E $: increasing and concave up.

Thus, the interval satisfying both conditions is $ A $ to $ B $.
✅ Answer: (A)

Question

Which one of these intervals could $g$ have an average rate of change of $-2$ and have rates of change that are changing at a rate of $-2$?

a. $[A, B]$
b. $[C, D]$
c. $[E, F]$
d. $[G, H]$

▶️ Answer/Explanation

Detailed solution

The average rate of change of $-2$ implies the slope of the secant line between the points is negative.
The rate of change is “changing at a rate of $-2$” means the second derivative $g”(x)$ is negative.
An interval where $g”(x) < 0$ corresponds to the graph being concave down.
Interval $[A, B]$ has a negative slope but is concave up.
Interval $[C, D]$ has a positive slope and is concave up.
Interval $[E, F]$ has a positive slope and is concave down.
Interval $[G, H]$ has a negative slope and is concave down.
Therefore, the only interval that satisfies both conditions is d.$[G, H]$.

Question

A wave machine produces a wave of $y$ height after $x$ amount of time according to the equation $y = 3.53 \sin(3\sqrt{x}/3) + 3$ for $x > 0$. At when the height first reaches $6$ ft, which choice best describes the rate of change, and the rate of change is changing?

a. The rate of change is increasing, and the rate of change is changing at an increasing rate.
b. The rate of change is decreasing, and the rate of change is changing at a decreasing rate.
c. The rate of change is increasing, and the rate of change is changing at a decreasing rate.
d. The rate of change is decreasing, and the rate of change is changing at an increasing rate.

▶️ Answer/Explanation

Detailed solution

The wave height is $y = 3.53 \sin(\sqrt{x}) + 3$.
To find when $y = 6$, we solve $3.53 \sin(\sqrt{x}) + 3 = 6$, giving $\sin(\sqrt{x}) \approx 0.85$.
The first derivative $y’ = \frac{3.53 \cos(\sqrt{x})}{2\sqrt{x}}$ represents the rate of change.
At the first instance $y=6$, $\sqrt{x} \approx 1.01$ rad, so $\cos(\sqrt{x}) > 0$, meaning the rate is positive but decreasing as it nears the peak.
The second derivative $y”$ represents how the rate is changing; here, it is negative.
The third derivative $y”’$ shows the rate of change of the rate of change is increasing (becoming less negative).
Therefore, the correct description is Option d.

Question

(A) The interval from $ A $ to $ B $

(B) The interval from $ B $ to $ C $

(D) The interval from $ D $ to $ E $

▶️ Answer/Explanation

Detailed solution

The problem asks for an interval where the function $ f $ is both increasing and concave down.
1. Increasing: The graph of $ f $ goes upwards (positive slope) on the intervals from $ A $ to $ B $ and from $ D $ to $ E $.
2. Concave Down: The graph curves downwards (like an inverted bowl or “frown”) on the interval from $ A $ to the inflection point $ C $.
3. Intersection: We need the interval that satisfies both conditions simultaneously.
4. The interval from $ A $ to $ B $ is within the region where the graph is rising and curving downwards.
5. The interval from $ D $ to $ E $ is increasing but concave up (curving upwards like a “smile”).
6. Therefore, the only interval that meets both criteria is from $ A $ to $ B $.

Correct Option: (A)

Question

Consider the piecewise function $W$ defined by $$W(t) = \begin{cases} 125 – 0.2t^2 & \text{for } 0 \le t < 20 \\ 445 – 20t & \text{for } 20 \le t \le 22 \end{cases}$$

Which of the following is a correct statement about the average rates of change of $W$?

(A) The average rate of change of $W$ for $0 \le t \le 5$ is less than the average rate of change of $W$ for $5 \le t \le 10$.

(B) The average rate of change of $W$ for $10 \le t \le 15$ is less than the average rate of change of $W$ for $5 \le t \le 10$.

(C) The average rate of change of $W$ for $15 \le t \le 20$ is greater than the average rate of change of $W$ for $10 \le t \le 15$.

(D) The average rate of change of $W$ for $21 \le t \le 22$ is greater than the average rate of change of $W$ for $20 \le t \le 21$.

▶️ Answer/Explanation

(B)

Detailed solution

To verify option (B), we calculate the function values at $t=5$, $t=10$, and $t=15$.
Using $W(t) = 125 – 0.2t^2$, we find $W(5) = 120$, $W(10) = 105$, and $W(15) = 80$.
The average rate of change for $5 \le t \le 10$ is $\frac{105 – 120}{10 – 5} = \frac{-15}{5} = -3$.
The average rate of change for $10 \le t \le 15$ is $\frac{80 – 105}{15 – 10} = \frac{-25}{5} = -5$.
Comparing the values, we see that $-5 < -3$.
Thus, the average rate of change for $10 \le t \le 15$ is less than that for $5 \le t \le 10$.

Question

The function ( p ) is given by ( p(x) = g(x + 1) – g(x) ). If ( p(x) = 2 ), which of the following statements must be true?

I. Because ( p ) is positive and constant, the graph of ( g ) always has positive slope.
II. Because ( p ) is positive and constant, the graph ( g ) is concave up
III. Because ( p ) is positive and constant, ( g ) is increasing

(A) I only
(B) II only
(C) III only
(D) I and III only

▶️ Answer/Explanation

Detailed solution

The condition $ g(x+1) – g(x) = 2 $ defines a constant average rate of change over an interval of 1, not the instantaneous slope (derivative).
Statement I is false: A function can oscillate while trending up (e.g., $ g(x) = 2x + \sin(2\pi x) $). Its slope $ g'(x) $ can be negative in certain intervals despite the net increase.
Statement II is false: A linear function like $ g(x) = 2x $ satisfies the condition (since $ 2(x+1) – 2x = 2 $) but has zero concavity, meaning it is not concave up.
Statement III is true (by elimination): Since I and II are false via counter-examples, III is the only plausible option, referring to the function’s net increasing trend.
Therefore, the correct choice is (C) III only.

Question

Given the graph of the function $y = f(x)$ shown above with points $a, b, c, d, e, f, g$, and $h$, which of the following pairs of points below has the greatest average rate of change?

(A) $a$ and $c$

(B) $b$ and $c$

(D) $f$ and $h$

▶️ Answer/Explanation

Detailed solution

The average rate of change between two points is equivalent to the slope of the line connecting them, calculated as $m = \frac{y_2 – y_1}{x_2 – x_1}$.
We are looking for the “greatest” rate, which implies the largest positive slope. Pairs $(a, c)$ and $(b, c)$ represent sections where the graph is decreasing, meaning their slopes are negative.
For option (C), we identify the coordinates from the grid: point $c$ is at $(-1, -4)$ and point $d$ is at $(1, 0)$.
Calculating the slope for pair $(c, d)$: $m = \frac{0 – (-4)}{1 – (-1)} = \frac{4}{2} = 2$.
For option (D), point $f$ is at $(4, -1)$ and $h$ is approximately at $(5.8, 1.8)$. The estimated slope is roughly $\frac{1.8 – (-1)}{5.8 – 4} \approx 1.55$.
Comparing the positive values, the slope of 2 is greater than $\approx 1.55$.
Therefore, the pair $c$ and $d$ has the greatest average rate of change.

Correct Option: (C)

Question

An object travels along a horizontal line with velocity given by the function $ v $. The graph of $ y = v(t) $, which consists of 4 line segments, is shown for $ 0 \le t \le 10 $. On which of the following intervals is the object’s velocity decreasing over the entire interval?

(A) $ 0 \le t \le 3 $

(B) $ 3 \le t \le 5 $

(D) $ 8 \le t \le 10 $

▶️ Answer/Explanation

Detailed solution

The velocity $ v(t) $ is decreasing when the acceleration $ a(t) $ is negative. On a velocity-time graph, this corresponds to intervals where the slope of the line is negative.

$\bullet$ For $ \mathbf{0 \le t \le 3} $: The graph slopes downward from $ v=0 $ to $ v=-2 $. The slope is negative ($ m = -2/3 $), so velocity is decreasing.
$\bullet$ For $ 3 \le t \le 5 $: The graph slopes upward. The slope is positive, so velocity is increasing.
$\bullet$ For $ 5 \le t \le 8 $: The graph is horizontal. The slope is zero, so velocity is constant.
$\bullet$ For $ \mathbf{8 \le t \le 10} $: The graph slopes downward from $ v=2 $ to $ v=0 $. The slope is negative ($ m = -1 $), so velocity is decreasing.

Conclusion: The velocity is decreasing on the intervals $ 0 \le t \le 3 $ and $ 8 \le t \le 10 $. Both options (A) and (D) are mathematically correct intervals for decreasing velocity.

Question

The figure shows a sphere that can expand. As it expands, the volume inside the sphere and the radius of the sphere both increase. In particular, the radius increases at a decreasing rate with respect to the volume.

Which of the following graphs could depict this situation, where volume, in cubic centimeters, is the independent variable and radius, in centimeters, is the dependent variable?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The relationship between volume $V$ and radius $r$ is given by $V = \frac{4}{3}\pi r^3$.
Solving for the dependent variable radius, we get $r = \sqrt[3]{\frac{3V}{4\pi}}$.
Since $r$ is proportional to $V^{1/3}$, the radius increases as volume increases.
The rate of change is $\frac{dr}{dV} = \frac{1}{3} \cdot (\frac{3}{4\pi})^{1/3} \cdot V^{-2/3}$.
As $V$ increases, $V^{-2/3}$ decreases, meaning the rate of increase is slowing down.
Graph B shows an increasing function with a decreasing slope (concave down).
Therefore, the correct graph is (B).

Question

An object travels along a horizontal line with velocity given by the function $v$. The graph of $y = v(t)$, which consists of four line segments, is shown for $0 \le t \le 10$. On which of the following intervals is the object’s velocity decreasing over the entire interval?

(A) $0 < t < 4$
(B) $5 < t < 8$
(C) $8 < t < 10$ only
(D) $0 < t < 3$ and $8 < t < 10$

▶️ Answer/Explanation

Detailed solution

Velocity is decreasing where the graph of $v(t)$ has a negative slope.
On the interval $0 < t < 3$, the graph moves from $y = 0$ down to $y = -2$, showing a decrease.
On the interval $3 < t < 5$, the graph moves from $y = -2$ up to $y = 2$, showing an increase.
On the interval $5 < t < 8$, the velocity is constant at $v(t) = 2$.
On the interval $8 < t < 10$, the graph moves from $y = 2$ down to $y = 0$, showing a decrease.
Therefore, the velocity is decreasing on the intervals $0 < t < 3$ and $8 < t < 10$.
The correct option is (D).

Question

Consider the piecewise function $W$ defined by: $$W(t) = \begin{cases} 125 – 0.2t^2 & \text{for } 0 \le t < 20 \\ 445 – 20t & \text{for } 20 \le t \le 22 \end{cases}$$ Which of the following is a correct statement about the average rates of change of $W$?

(A) The average rate of change of $W$ for $0 \le t \le 5$ is less than the average rate of change of $W$ for $5 \le t \le 10$.
(B) The average rate of change of $W$ for $10 \le t \le 15$ is less than the average rate of change of $W$ for $5 \le t \le 10$.
(C) The average rate of change of $W$ for $15 \le t \le 20$ is greater than the average rate of change of $W$ for $10 \le t \le 15$.
(D) The average rate of change of $W$ for $21 \le t \le 22$ is less than the average rate of change of $W$ for $20 \le t \le 21$.

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
For $0 \le t < 20$, $W(t) = 125 – 0.2t^2$ is a downward-opening parabola.
The average rate of change on $[a, b]$ is $\frac{W(b) – W(a)}{b – a}$.
On $[5, 10]$, the rate is $\frac{(125 – 20) – (125 – 5)}{10 – 5} = \frac{105 – 120}{5} = -3$.
On $[10, 15]$, the rate is $\frac{(125 – 45) – (125 – 20)}{15 – 10} = \frac{80 – 105}{5} = -5$.
Since $-5 < -3$, the average rate of change for $10 \le t \le 15$ is less than for $5 \le t \le 10$.
For linear parts (Option D), the rate is constant at $-20$, so they are equal, not less.

Question

Which of the following statements about the graph of $ f $ is correct?

(A) On the interval $ 1 < x < 3 $, $ f $ is increasing at an increasing rate.

(B) On the interval $ 1 < x < 3 $, $ f $ is increasing at a decreasing rate.

(D) On the interval $ 1 < x < 3 $, $ f $ is decreasing at a decreasing rate.

▶️ Answer/Explanation

Detailed solution

Step 1: Observe the direction of the curve on the interval $ 1 < x < 3 $. As $ x $ increases from 1 to 3, the value of $ y $ goes down. Therefore, the function $ f $ is decreasing. This eliminates options (A) and (B).

Step 2: Analyze the rate of change (concavity). The “rate” refers to the steepness or magnitude of the slope.

Step 3: On the interval $ 1 < x < 3 $, the graph is part of the downward slope coming off the local maximum at $ x=0 $. The curve is getting steeper as it goes down (it forms an inverted bowl shape, which is concave down).

Step 4: Since the slope is becoming more negative (steeper), the rate of decrease is increasing.

Conclusion: The function is decreasing, and it is doing so at an increasing rate. Thus, statement (C) is correct.

Question

The figure shows the graph of a function $h$ in the $xy$-plane with four labeled points. It is known that a relative maximum of $h$ occurs at $C$. For which of the following pairs of points is the average rate of change of $h$ the least?

(A) between points $A$ and $C$

(B) between points $B$ and $C$

(D) between points $C$ and $D$

▶️ Answer/Explanation

Detailed solution

The average rate of change is the slope of the secant line connecting two points: $m = \frac{h(x_2) – h(x_1)}{x_2 – x_1}$.
Between $A(0.5, 2)$ and $C(6.8, 1.3)$, the slope is slightly negative.
Between $B(4, -1.2)$ and $C(6.8, 1.3)$, the slope is positive as the function increases.
Between $B(4, -1.2)$ and $D(8.5, -2.8)$, the slope is $\frac{-2.8 – (-1.2)}{8.5 – 4} = \frac{-1.6}{4.5} \approx -0.35$.
Between $C(6.8, 1.3)$ and $D(8.5, -2.8)$, the slope is $\frac{-2.8 – 1.3}{8.5 – 6.8} = \frac{-4.1}{1.7} \approx -2.41$.
The value $-2.41$ is the most negative, representing the least average rate of change.
Therefore, the average rate of change is least between points $C$ and $D$.

Question

Let $g$ be the piecewise function defined by $$g(x) = \begin{cases} 1 – x^2 & \text{for } -5 \le x < 2 \\ 3 + 2x & \text{for } 2 \le x \le 9 \end{cases}$$ What is the average rate of change of $g$ over the closed interval $[-2, 4]$?

(A) $-2$

(B) $1$

(D) $\frac{7}{3}$

▶️ Answer/Explanation

Detailed solution

The average rate of change on $[a, b]$ is given by the formula $\frac{g(b) – g(a)}{b – a}$.
Identify the interval boundaries: $a = -2$ and $b = 4$.
Calculate $g(-2)$ using the first piece: $g(-2) = 1 – (-2)^2 = 1 – 4 = -3$.
Calculate $g(4)$ using the second piece: $g(4) = 3 + 2(4) = 3 + 8 = 11$.
Substitute values into the formula: $\text{Rate} = \frac{11 – (-3)}{4 – (-2)}$.
Simplify the expression: $\text{Rate} = \frac{14}{6} = \frac{7}{3}$.
The correct option is (D).

Question

The table describes rates of change of a function $f$ for selected intervals of $x$. The function $f$ is defined for $0 \le x \le 4$. On which of the following intervals is the graph of $f$ concave down?

(A) $0 < x < 1$
(B) $1 < x < 2$
(C) $2 < x < 3$
(D) $3 < x < 4$

▶️ Answer/Explanation

Detailed solution

The concavity of a function $f$ is determined by the behavior of its rate of change, $f’$.
A graph is concave down on an interval if its rate of change is decreasing.
In the interval $0 < x < 1$, the rate of change is increasing, so $f$ is concave up.
In the interval $1 < x < 2$, the rate of change is constant, so there is no concavity.
In the interval $2 < x < 3$, the table explicitly states the rate of change is decreasing.
In the interval $3 < x < 4$, the rate of change is constant, so there is no concavity.
Therefore, the graph of $f$ is concave down only on the interval $2 < x < 3$.
Correct Option: (C)

Question

Two drones are flying over a given area, and their heights above the ground are changing. The table gives the change in height, in feet, for the drones over successive $6$-second intervals. Which of the following is true about the average rates of change for drone A and drone B over the time interval from $t = 0$ seconds to $t = 30$ seconds?

▶️ Answer/Explanation

Detailed solution

The total change in height for Drone A is $17 + (-4) + 11 + (-5) + (-3) = 16$ feet.
The total change in height for Drone B is $5 + 3 + 3 + 2 + 3 = 16$ feet.
Average rate of change is calculated as $\frac{\text{Total Change in Height}}{\text{Total Time}}$.
For Drone A, the average rate of change is $\frac{16}{30}$ feet per second.
For Drone B, the average rate of change is $\frac{16}{30}$ feet per second.
Since both total changes and total times are identical, the average rates are equal.
Therefore, the correct choice is (A).

Question

Consider the piecewise function $W$ defined by $W(t) = \begin{cases} 125 – 0.2t^2 & \text{for } 0 \le t < 20 \\ 445 – 20t & \text{for } 20 \le t \le 22 \end{cases}$ Which of the following is a correct statement about the average rates of change of $W$?

(A) The average rate of change of $W$ for $0 \le t \le 5$ is less than the average rate of change of $W$ for $5 \le t \le 10$.

(B) The average rate of change of $W$ for $10 \le t \le 15$ is less than the average rate of change of $W$ for $5 \le t \le 10$.

(C) The average rate of change of $W$ for $15 \le t \le 20$ is greater than the average rate of change of $W$ for $10 \le t \le 15$.

(D) The average rate of change of $W$ for $21 \le t \le 22$ is less than the average rate of change of $W$ for $20 \le t \le 21$.

▶️ Answer/Explanation

Detailed solution

The average rate of change on $[a, b]$ is $\frac{W(b) – W(a)}{b – a}$.
For $0 \le t < 20$, $W(t) = 125 – 0.2t^2$, which is a downward-opening parabola.
The rate of change on $[5, 10]$ is $\frac{W(10)-W(5)}{5} = \frac{105-120}{5} = -3$.
The rate of change on $[10, 15]$ is $\frac{W(15)-W(10)}{5} = \frac{80-105}{5} = -5$.
Since $-5 < -3$, statement (B) is correct: the rate on $[10, 15]$ is less than on $[5, 10]$.
For $20 \le t \le 22$, $W(t)$ is linear with a constant slope of $-20$.
Thus, the rates on $[20, 21]$ and $[21, 22]$ are equal, making (D) incorrect.

Question

A toy car travels around a circular track as shown. The distance from the wall is graphed against time as the car completes three full revolutions without stopping. Which graph correctly models this relationship?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Correct Answer: (A)

Explanation:
As the car moves with constant speed around a circular track, its horizontal position relative to the wall varies sinusoidally with time. The distance from a vertical wall depends on the horizontal coordinate of circular motion, which can be modeled as: \[ x(t) = r \cos(\omega t) \] Since distance must be non-negative, the graph represents a shifted cosine wave. Over three revolutions, the pattern repeats three times smoothly. Only option (A) shows a smooth periodic curve that stays non-negative and repeats three times.

Detailed solution

1. Circular motion with constant speed gives horizontal position $ x(t) = r\cos(\omega t) $.
2. Distance from the wall depends on this horizontal coordinate.
3. The function is periodic with period $ T = \dfrac{2\pi}{\omega} $.
4. Three revolutions produce three identical cycles.
5. Distance cannot be negative, so the curve stays above the time axis.
6. Only graph (A) shows a smooth repeating periodic wave three times.

Question

The graph of the function $f$ is given for $-3 \le x \le 6$. Which of the following statements about the rate of change of $f$ over the interval $2 < x < 6$ is true?

(A) The rate of change is positive.
(B) The rate of change is negative.
(C) The rate of change is increasing.
(D) The rate of change is decreasing.

▶️ Answer/Explanation

Detailed solution

The rate of change of a function is represented by the slope of the tangent line to the graph.
On the interval $2 < x < 4$, the function is increasing, so the rate of change is positive.
On the interval $4 < x < 6$, the function is decreasing, so the rate of change is negative.
Throughout the entire interval $2 < x < 6$, the graph is concave down (it curves downward).
For a concave down graph, the slope of the tangent line continuously decreases as $x$ increases.
The slope starts as a large positive value at $x = 2$, becomes $0$ at $x \approx 4$, and becomes negative for $x > 4$.
Therefore, the correct statement is that the rate of change is decreasing.
Correct Option: (D)

Question

The figure shows a swimming pool filled with water. A pump is used to remove water from the pool until the pool is empty. When the pump is running, the rate at which the volume of water in the pool decreases is constant.

During the first $2$ hours, the pump works slower than usual due to a broken piece. Then the pump stops working. The broken piece is replaced, and the pump works at its usual rate until the pool is completely emptied of water. The entire process of emptying the pool takes $6$ hours.

Which of the following graphs could depict this situation, where time, in hours, is the independent variable, and the volume of water in the pool, in gallons, is the dependent variable?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
During the first $t = 2$ hours, the volume decreases slowly, represented by a shallow negative slope.
When the pump stops working, the volume remains constant, represented by a horizontal line segment.
After the repair, the pump works at a “usual” (faster) rate, represented by a steeper negative slope.
The graph must intersect the $x$-axis (volume $= 0$) exactly at $t = 6$ hours.
Graph (A) is incorrect because it ends at $t = 10$ hours.
Graph (C) is incorrect because the initial slope is steeper than the final slope.
Graph (D) is incorrect because the pump remains stopped until $t = 5$ and the process ends at $t = 6$.

AP Precalculus -1.2 Rates of Change- MCQ Exam Style Questions - Effective Fall 2023