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)

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