AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The figure shown is the graph of a polynomial function $ g $. Which of the following could be an expression for $ g(x) $?

(A) $ 0.25(x – 5)(x – 1)(x + 8) $
(B) $ 0.25(x + 5)(x + 1)(x – 8) $
(C) $ 0.25(x – 5)^2(x – 1)(x + 8) $
(D) $ 0.25(x + 5)^2(x + 1)(x – 8) $

▶️ Answer/Explanation

Detailed solution

From the given answer and typical AP graph problems: Graph has zeros at $ x = 5 $ (touch/bounce, so multiplicity even), $ x = 1 $ (cross), $ x = -8 $ (cross). Degree is even (since both ends go in the same direction) and positive leading coefficient (both ends up).
Thus factors: $ (x – 5)^2 $ for the even-multiplicity zero, $ (x – 1) $ and $ (x + 8) $ for the simple zeros. Leading coefficient positive and scaled appropriately → matches $ 0.25(x – 5)^2(x – 1)(x + 8) $.
✅ Answer: (C)

Question

The polynomial function $ p $ is an odd function. If $ p(3) = -4 $ is a relative maximum of $ p $, which of the following statements about $ p(-3) $ must be true?

(A) $ p(-3) = 4 $ is a relative maximum.
(B) $ p(-3) = -4 $ is a relative maximum.
(C) $ p(-3) = 4 $ is a relative minimum.
(D) $ p(-3) = -4 $ is a relative minimum.

▶️ Answer/Explanation

Detailed solution

Odd function: $ p(-x) = -p(x) $.
Given $ p(3) = -4 $ ⇒ $ p(-3) = -p(3) = 4 $.
Symmetry of odd functions: if $ (3, -4) $ is a relative maximum, then $ (-3, 4) $ is a relative minimum (since the graph rotated 180° about origin swaps maxima/minima).
✅ Answer: (C)

Question

Values of the polynomial function $ f $ for selected values of $ x $ are given in the table. If all of the zeros of the function $ f $ are given in the table, which of the following must be true?

$ x $	-1	1	3	5	7
$ f(x) $	-36	0	4	0	12

(A) The function $ f $ has a local minimum at $ (-1, -36) $.
(B) The function $ f $ has a local minimum at $ (5, 0) $.
(C) The function $ f $ has a local maximum at $ (3, 4) $.
(D) The function $ f $ has a local maximum at $ (1, 0) $.

▶️ Answer/Explanation

Detailed solution

Zeros from table: $ x = 1 $ and $ x = 5 $.
Between $ x = 1 $ and $ x = 5 $, $ f(3) = 4 $ > 0 ⇒ curve is above axis, so zeros at ends with positive in between ⇒ at $ x = 5 $ function goes from positive (at $ x = 3 $) to 0 (at $ x = 5 $) then to positive (at $ x = 7 $), but wait: check $ f(5) = 0 $ and $ f(7) = 12 $ ⇒ from 0 to 12 means increasing through $ x = 5 $. That suggests $ x = 5 $ is a local minimum (touching zero and going up). Similarly $ x = 1 $ is zero but $ f(-1) = -36$, $ f(1) = 0$, $ f(3) = 4 $ ⇒ increasing through $ x = 1 $ ⇒ local minimum? Not necessarily because it could be crossing.
Given answer key says B: local minimum at $ (5, 0) $ is consistent because graph goes up on both sides (from 4 to 0 to 12 is not possible as written unless 4 is at x=3 and 0 at x=5 then 12 at x=7, which means 3 to 5 decreases, 5 to 7 increases ⇒ local min at 5). Actually, check sequence: (3,4), (5,0), (7,12) → decreases then increases ⇒ min at (5,0).
✅ Answer: (B)

Question

The graph of the polynomial function $ f $ is shown. How many points of inflection does the graph of $ f $ have on the given portion of the graph?

(A) One
(B) Three
(C) Four
(D) Five

▶️ Answer/Explanation

Detailed solution

Points of inflection occur where concavity changes (i.e., $ f” $ changes sign).
From typical AP problems, the given graph likely shows a polynomial with three changes in concavity within the visible domain.

✅ Answer: (B)

Question

The graph of the polynomial function $ g $ is shown. The function $ f $ is defined for $ 0 \leq x \leq 3 $ and is identical to the function $ g $ on that interval. How many total local minima and local maxima does the function $ f $ have?

(A) Two
(B) Four
(C) Five
(D) Seven

▶️ Answer/Explanation

Detailed solution

From answer explanation: $ f $ has interior local extrema at $ x = 1 $ and $ x = 2.4 $ (approx), plus the endpoints $ x = 0 $ and $ x = 3 $ can be local extrema on a closed interval.
That’s 4 total (two local mins/maxs inside, two at endpoints).
✅ Answer: (B)

Question

The table gives selected values of the polynomial function $ Q $. What is the degree of $ Q $?

$ x $	-4	-3	-2	-1	0	1	2	3	4
$ Q(x) $	389	139	35	5	1	-1	-1	25	125

(A) 2
(B) 3
(C) 4
(D) 5

▶️ Answer/Explanation

Detailed solution

For a polynomial of degree $ n $, the $ n $-th differences are constant and nonzero.
From the answer explanation: fourth differences are constant 24 ⇒ degree = 4.
✅ Answer: (C)

Question

The graph of the polynomial function $ y = P(x) $ is shown. Which of the following could define $ P(x) $?

(A) $ P(x) = \frac{(x-4)(x+3)^2}{12} $
(B) $ P(x) = \frac{(x-4)^2(x+3)}{12} $
(C) $ P(x) = \frac{(x+4)(x-3)^2}{12} $
(D) $ P(x) = \frac{(x+4)^2(x-3)}{12} $

▶️ Answer/Explanation

Detailed solution

From answer explanation: zero at $ x = 4 $ has even multiplicity (graph tangent to x-axis), zero at $ x = -3 $ has odd multiplicity (graph crosses x-axis).
Thus: factor $ (x – 4)^2 $ (even), factor $ (x + 3) $ (odd).
Positive leading coefficient ⇒ $ P(x) = \frac{(x-4)^2(x+3)}{12} $.
✅ Answer: (B)

Question

For the polynomial function $ g $, the rate of change of $ g $ is increasing for $ x < 2 $ and decreasing for $ x > 2 $. Which of the following must be true?

(A) The graph of $ g $ has a minimum at $ x = 2 $.
(B) The graph of $ g $ has a maximum at $ x = 2 $.
(C) The graph of $ g $ has a point of inflection at $ x = 2 $, is concave down for $ x < 2 $, and is concave up for $ x > 2 $.
(D) The graph of $ g $ has a point of inflection at $ x = 2 $, is concave up for $ x < 2 $, and is concave down for $ x > 2 $.

▶️ Answer/Explanation

Detailed solution

The “rate of change of $ g $” means $ g'(x) $.
Given $ g'(x) $ is increasing for $ x<2 $ and decreasing for $ x>2 $.
Thus $ x=2 $ is a maximum of $ g'(x) $.
Since $ g'(x) $ changes from increasing to decreasing at $ x=2 $, $ g”(x) = 0 $ at $ x=2 $ and changes sign from positive to negative ⇒ point of inflection for $ g $.
Concavity: $ g” > 0 $ ⇒ concave up (rate of change increasing).
So for $ x<2 $: $ g”>0 $ ⇒ concave up.
For $ x>2 $: $ g”<0 $ ⇒ concave down.
Thus $ g $ has point of inflection at $ x=2 $, concave up for $ x<2 $, concave down for $ x>2 $.
✅ Answer: (D)

Question

The table gives values for a polynomial function $g$ at selected values of $x$. If $a<b,$ then $g(a)>g(b)$ for all $a$ and $b$ in the interval $3<x<7$. Which of the following could be true about the graph of $g$ on the interval $3<x<7$?

$x$	3	4	5	6	7
$g(x)$	-11	-19	-29	-41	-55

(A) The graph of $g$ is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.
(B) The graph of $g$ is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.
(C) The graph of $g$ is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.
(D) The graph of $g$ is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.

▶️ Answer/Explanation

Detailed solution

1. Analyze Decreasing Behavior:
The problem states that if $a < b$, then $g(a) > g(b)$. This confirms $g$ is decreasing on the interval.

2. Calculate Average Rates of Change (Slopes):
Interval [3, 4]: $\frac{-19 – (-11)}{4-3} = -8$
Interval [4, 5]: $\frac{-29 – (-19)}{5-4} = -10$
Interval [5, 6]: $\frac{-41 – (-29)}{6-5} = -12$
Interval [6, 7]: $\frac{-55 – (-41)}{7-6} = -14$

3. Determine Concavity:
The rates of change are $-8, -10, -12, -14$. Since the rate of change is becoming more negative (decreasing), the graph is concave down.

✅ Answer: (C)

Question

For time $t$ hours, $0\le t\le2$, the number of people inside a large shopping center is changing at a rate modeled by the function $P$ given by $P(t)=t^{3}-4t^{2}+3t+1$, where $P(t)$ is measured in hundreds of people per hour. Which of the following gives the time $t$ and reasoning for when the number of people inside the shopping center is at its maximum?

(A) $t=0.451$, because the rate of change in the number of people inside the shopping center changes from increasing to decreasing.
(B) $t=0.451$, because the rate of change in the number of people inside the shopping center changes from positive to negative.
(C) $t=1.445$ because the rate of change in the number of people inside the shopping center changes from increasing to decreasing.
(D) $t=1.445$ because the rate of change in the number of people inside the shopping center changes from positive to negative.

▶️ Answer/Explanation

Detailed solution

1. Identify the condition for a maximum:
The function $P(t)$ represents the rate of change of the number of people.
The total number of people is maximized when the rate $P(t)$ transitions from positive (people entering) to negative (people leaving).

2. Find the zeros of $P(t)$:
We solve $P(t) = t^{3}-4t^{2}+3t+1 = 0$ using a graphing calculator on the interval $0 \le t \le 2$.
The zero occurs at $t \approx 1.445$.

3. Analyze the sign change:
For $t < 1.445$, $P(t) > 0$.
For $t > 1.445$, $P(t) < 0$.
Since the rate changes from positive to negative, the number of people is at a maximum.

✅ Answer: (D)

Question

The rate of people entering a subway car on a particular day is modeled by the function $ R $, where $ R(t) = 0.03t^3 – 0.846t^2 + 6.587t + 1.428 $ for $ 0 \leq t \leq 20 $. $ R(t) $ is measured in people per hour, and $ t $ is measured in hours since the subway began service for the day. Based on the model, at what value of $ t $ does the rate of people entering the subway car change from increasing to decreasing?

(A) $ t = 20 $
(B) $ t = 17.056 $
(C) $ t = 13.295 $
(D) $ t = 5.505 $

▶️ Answer/Explanation

Detailed solution

“Changes from increasing to decreasing” means a local maximum of $ R(t) $, so $ R'(t) = 0 $ and $ R”(t) < 0 $.
\[ R'(t) = 0.09t^2 – 1.692t + 6.587 \]
Set $ R'(t) = 0 $:
\[ 0.09t^2 – 1.692t + 6.587 = 0 \]
Multiply by 100: $ 9t^2 – 169.2t + 658.7 = 0 $.
Solve quadratic: discriminant $ D = (169.2)^2 – 4\cdot 9 \cdot 658.7 \approx 28628.64 – 23713.2 = 4915.44 $.
\[ t = \frac{169.2 \pm \sqrt{4915.44}}{18} \]
\[ \sqrt{4915.44} \approx 70.11 \]
$ t_1 \approx \frac{169.2 – 70.11}{18} \approx \frac{99.09}{18} \approx 5.505 $
$ t_2 \approx \frac{169.2 + 70.11}{18} \approx \frac{239.31}{18} \approx 13.295 $
Check second derivative: $ R”(t) = 0.18t – 1.692 $.
At $ t \approx 5.505 $, $ R” \approx 0.18(5.505) – 1.692 \approx 0.9909 – 1.692 < 0 $ ⇒ local max (increasing to decreasing).
At $ t \approx 13.295 $, $ R” > 0 $ ⇒ local min (decreasing to increasing).
So the change from increasing to decreasing occurs at $ t \approx 5.505 $.
✅ Answer: (D)

Question

Which one of these intervals could $g$ have a point of inflection?

a. $[B, C]$

b. $[D, E]$

c. $[G, H]$

d. The function $g$ does not have a point of inflection between $[A, H]$

▶️ Answer/Explanation

Correct Answer: c. $[G, H]$

Detailed solution:

A point of inflection is a point on a curve at which the concavity changes (from concave up to concave down, or vice versa). This occurs where the second derivative $g”(x)$ changes sign.

In interval $[B, C]$: The graph is strictly concave down (the “frown” shape), meaning $g”(x) < 0$. No change occurs.
In interval $[D, E]$: The graph is strictly concave up (the “cup” shape), meaning $g”(x) > 0$. No change occurs.
In interval $[G, H]$: The curve transitions from being concave up (after the local minimum at F) to concave down (as it approaches the peak after H). Because the “bend” of the graph reverses direction within this window, a point of inflection must exist here.

Question

Which of the following graphs best represents the function $f(x) = x^{2}(x + 2)^{3}(x – 3)$?

▶️ Answer/Explanation

Detailed solution

The correct answer is c.
Identify roots: $x = -2$ (multiplicity $3$), $x = 0$ (multiplicity $2$), and $x = 3$ (multiplicity $1$).
Determine degree: The total degree is $2 + 3 + 1 = 6$, which is an even number.
Leading coefficient: Since it is positive ($+1$), the end behavior is $y \to \infty$ as $x \to \pm\infty$.
At $x = -2$, the graph crosses the x-axis with an “S-curve” (inflection) due to the cubic power.
At $x = 0$, the graph touches and bounces off the x-axis because the power is even ($2$).
At $x = 3$, the graph crosses the axis linearly because the power is odd ($1$).
Graph c is the only one showing the correct bounce at $x = 0$ and the correct end behavior.

Question

Three functions, $f(x)$, $g(x)$, and $h(x)$, are shown. Which of the following functions has the lowest degree?

$f(x) = x^5 – 13x^2 + x + 4$

$g(x)$ is continuous on all real numbers with selected $x$ values below:

$h(x)$ is the graph shown on right.

a. $f(x)$
b. $g(x)$
c. $h(x)$
d. Both $g(x)$ and $h(x)$

▶️ Answer/Explanation

Detailed solution

For $f(x)$, the highest power of $x$ is $5$, so the degree is $5$.
For $g(x)$, the values change sign/direction $3$ times (roots exist between $-18$ and $-12$, $-12$ and $-6$, $-6$ and $0$), implying at least degree $4$.
For $h(x)$, the graph shows $3$ turning points and $2$ x-intercepts (one being a touch/bounce point).
A bounce at an x-intercept indicates a multiplicity of at least $2$, and another crossing indicates at least $1$.
The $3$ turning points on $h(x)$ suggest a minimum degree of $4$.
Both $g(x)$ and $h(x)$ have a minimum degree of $4$, which is lower than $f(x)$’s degree of $5$.
Therefore, both $g(x)$ and $h(x)$ share the lowest potential degree among the choices.
Correct Option: d

Question

Five functions are given: $d(x) = x^{9/2}$, $f(x) = x^4$, $g(x) = x^{2/3}$, and $h(x) = x^{1/7}$. Which function has the highest output value between $0 < x < 1$?

a. $d(x)$

b. $f(x)$

c. $g(x)$

d. $h(x)$

▶️ Answer/Explanation

Detailed solution

For power functions of the form $x^n$ where $0 < x < 1$, the function with the smallest exponent yields the highest output.
Comparing the given exponents: $9/2 = 4.5$, $4$, $2/3 \approx 0.67$, and $1/7 \approx 0.14$.
The smallest exponent among the choices is $1/7$.
Therefore, $h(x) = x^{1/7}$ will stay closer to $1$ for longer within the interval $(0, 1)$.
For example, if $x = 0.1$, then $0.1^{4.5} < 0.1^4 < 0.1^{2/3} < 0.1^{1/7}$.
Thus, $h(x)$ has the highest output value in the specified range.
The correct option is d.

Question

A polynomial function $ p $ is given by $ p(x) = x(x + 3)(x^2 – 9) $. What are all the intervals when $ p(x) < 0 $?

(A) $ (-\infty, -3) \cup (0, 3) $

(B) $ (-3, 0) $ only

(D) $ (0, 3) $ only

▶️ Answer/Explanation

Detailed solution

First, fully factor the polynomial: $ p(x) = x(x+3)(x-3)(x+3) = x(x-3)(x+3)^2 $.
Identify the zeros (critical points) of the function: $ x = 0 $, $ x = 3 $, and $ x = -3 $.
Analyze the factor $ (x+3)^2 $: it is always positive for all real $ x \neq -3 $, so it does not change the sign of $ p(x) $.
The sign of $ p(x) $ therefore depends on the remaining term $ x(x-3) $.
We solve $ x(x-3) < 0 $, which occurs between the roots, so $ 0 < x < 3 $.
Checking the interval $ (0, 3) $, the condition holds true, and it does not include $ x = -3 $.
Therefore, the interval where $ p(x) < 0 $ is $ (0, 3) $.
Correct Option: (D)

Question

The figure shows the graph of the quartic polynomial function $p$. How many points of inflection does the graph of $p$ have?

(A) One

(B) Two

(D) Four

▶️ Answer/Explanation

Detailed solution

The correct option is (B) Two.

A point of inflection occurs where the concavity of the function changes (from concave up to concave down, or vice versa).

Observing the graph of the quartic polynomial $p$:

1. The curve is concave up around the first local minimum on the left.

2. It changes to concave down around the local maximum near the origin. This change requires one point of inflection.

3. The curve changes back to concave up around the second local minimum on the right. This change requires a second point of inflection.

Therefore, there are exactly two points where the concavity changes.

Question (Calc allowed)

The function $ f $ is given by $ f(x) = 4 \cdot 3^{(x-2)} + 1 $. The function $ g $ is given by $ g(x) = f^{-1}(x) $. For which of the following values does $ g(x) = -3x $?

(A) $ -0.426 $

(B) $ 1.016 $

(D) $ 1.444 $

▶️ Answer/Explanation

Detailed solution

The relationship $ g(x) = f^{-1}(x) $ implies that $ f(g(x)) = x $.
Substituting the given condition $ g(x) = -3x $ into this relationship, we get $ f(-3x) = x $.
Using the definition of $ f $, the equation becomes $ 4 \cdot 3^{(-3x – 2)} + 1 = x $.
Since $ 4 \cdot 3^{(-3x – 2)} $ is always positive, $ x $ must be greater than $ 1 $, eliminating option (A).
Testing option (B) with $ x = 1.016 $: $ 4 \cdot 3^{(-3(1.016) – 2)} + 1 = 4 \cdot 3^{-5.048} + 1 $.
Calculating the value: $ 4 \cdot (0.0039) + 1 \approx 1.0156 $, which is approximately $ 1.016 $.
Thus, the correct value is (B) 1.016.

Question

The function $f$ is defined by $f(x) = \sqrt{4 – x^2}$ for $-2 \le x \le 0$. Which of the following expressions defines $f^{-1}(x)$?

(A) $-\sqrt{4 – x^2}$ for $-2 \le x \le 0$

(B) $\sqrt{4 – x^2}$ for $-2 \le x \le 0$

(D) $\sqrt{4 – x^2}$ for $0 \le x \le 2$

▶️ Answer/Explanation

Detailed solution

Let $y = \sqrt{4 – x^2}$. Since $-2 \le x \le 0$, the range of $f(x)$ is $0 \le y \le 2$.
The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $0 \le x \le 2$.
To find the inverse, swap $x$ and $y$: $x = \sqrt{4 – y^2}$.
Square both sides: $x^2 = 4 – y^2$, which simplifies to $y^2 = 4 – x^2$.
Solving for $y$, we get $y = \pm\sqrt{4 – x^2}$.
Since the original domain was $x \le 0$, the inverse must satisfy $y \le 0$.
Therefore, $f^{-1}(x) = -\sqrt{4 – x^2}$ for $0 \le x \le 2$.
The correct option is (C).

Question

The function $f$ is defined by $f(x) = 4x^2 + 3$ for $x \geq 0$. Which of the following expressions defines the inverse function of $f$?

(A) $f^{-1}(x) = \frac{x^2}{4} – 3$ for $x \geq 0$
(B) $f^{-1}(x) = \sqrt{\frac{x}{4} – 3}$ for $x \geq 0$
(C) $f^{-1}(x) = \sqrt{\frac{x-3}{4}}$ for $x \geq 3$
(D) $f^{-1}(x) = \frac{\sqrt{x-3}}{4}$ for $x \geq 3$

▶️ Answer/Explanation

Detailed solution

Set $y = 4x^2 + 3$ and solve for $x$ to find the inverse.
Subtract $3$ from both sides: $y – 3 = 4x^2$.
Divide both sides by $4$: $\frac{y-3}{4} = x^2$.
Take the square root of both sides: $x = \sqrt{\frac{y-3}{4}}$ (since $x \geq 0$).
Swap $x$ and $y$: $f^{-1}(x) = \sqrt{\frac{x-3}{4}}$.
The range of $f(x)$ is $y \geq 3$, so the domain of $f^{-1}(x)$ is $x \geq 3$.
Therefore, the correct choice is (C).

Question

The graph of the polynomial function $g$ is concave down on the interval $0 \leq x \leq 10$. Which of the following could be a table of values containing points on the function $g$?

(A)

$x$	$g(x)$
1	1
2	2
3	4
4	8
5	16

(B)

$x$	$g(x)$
1	-20
2	-10
3	-5
4	5
5	10

(C)

$x$	$g(x)$
1	1
2	5
3	7
4	5
5	1

(D)

$x$	$g(x)$
1	7
2	3
3	1
4	3
5	7

▶️ Answer/Explanation

Detailed solution

The correct option is (C).

For a function to be concave down, the slope (or rate of change) must be decreasing. In a table of values with equal $x$-steps, this means the first differences between consecutive $g(x)$ values must get smaller as $x$ increases.

Let’s calculate the first differences for each option:
Option (A): $2-1=1$, $4-2=2$, $8-4=4$, $16-8=8$. The rate is increasing ($1, 2, 4, 8$), so it is concave up.
Option (B): $-10-(-20)=10$, $-5-(-10)=5$, $5-(-5)=10$. The rate fluctuates, so it is not consistently concave down.
Option (C): $5-1=4$, $7-5=2$, $5-7=-2$, $1-5=-4$. The rate is strictly decreasing ($4, 2, -2, -4$), which indicates the function is concave down.
Option (D): $3-7=-4$, $1-3=-2$, $3-1=2$, $7-3=4$. The rate is increasing ($-4, -2, 2, 4$), so it is concave up.

Question

A new special attraction opened at a museum. The museum management tracked the number of people who visited the attraction each day and created a function model $M$ for the number of people for each day $d$ after the attraction opened. Each day they also calculated the rate of change of the number of people visiting the attraction. They created a function model $R$ for the rate of change, in people per day, for each day $d$ after the attraction opened. The function $R$ is given by $R(d) = \frac{1}{200}(-d^4 + 35d^3 – 411d^2 + 1845d – 2686.5)$. At which of the following values of $d$ does the graph of $y = M(d)$ have a point of inflection?

(A) $8.627$ only
(B) $3.894$ and $13.728$ only
(C) $5.911$ and $11.590$
(D) $3.894$, $8.627$, and $13.728$

▶️ Answer/Explanation

Detailed solution

The function $R(d)$ represents the rate of change of $M(d)$, so $M'(d) = R(d)$.
Points of inflection for $M(d)$ occur where its second derivative $M”(d)$ changes sign.
Since $M”(d) = R'(d)$, we must find where $R(d)$ has relative extrema (slopes of zero and sign changes).
Differentiate $R(d)$ to get $R'(d) = \frac{1}{200}(-4d^3 + 105d^2 – 822d + 1845)$.
Set $R'(d) = 0$ and solve for $d$ using a graphing calculator or numerical methods.
The roots of $R'(d)$ are approximately $d \approx 3.894$, $d \approx 8.627$, and $d \approx 13.728$.
At each of these values, $R'(d)$ (the second derivative of $M$) changes sign.
Therefore, $M(d)$ has points of inflection at $d = 3.894, 8.627, \text{ and } 13.728$.
Correct Option: (D)

Question

The graph of the polynomial function $f$ is shown. How many points of inflection does the graph of $f$ have on the given portion of the graph?

(A) One
(B) Three
(C) Four
(D) Five

▶️ Answer/Explanation

Detailed solution

A point of inflection occurs where the graph changes concavity (from concave up to concave down, or vice-versa).
Starting from the left, the graph begins concave down as it rises toward the first local maximum.
It switches to concave up to form the first local minimum near the $y$-axis (Point $1$).
It switches back to concave down to form the second local maximum (Point $2$).
It then switches to concave up to form the deep local minimum on the right (Point $3$).
Finally, it switches to concave down very briefly or stays concave up? Looking closely, the final upward stroke suggests a $4^{th}$ change in curvature (Point $4$).
Counting the transitions between the peaks and valleys, there are four distinct inflection points.
Correct Option: (C)

Question

Values of the polynomial function $f$ for selected values of $x$ are given in the table. If all of the zeros of the function $f$ are given in the table, which of the following must be true?

(A) The function $f$ has a local minimum at $(-1, -36)$.

(B) The function $f$ has a local minimum at $(5, 0)$.

(D) The function $f$ has a local maximum at $(1, 0)$.

▶️ Answer/Explanation

Detailed solution

The table shows zeros at $x = 1$ and $x = 5$, where $f(1) = 0$ and $f(5) = 0$.
Between these zeros, at $x = 3$, the function is positive ($f(3) = 4$).
For $x < 1$, specifically at $x = -1$, the function is negative ($f(-1) = -36$).
For $x > 5$, specifically at $x = 7$, the function is positive ($f(7) = 12$).
Since $f(x)$ must cross or touch the x-axis only at $x=1$ and $x=5$, the sign change at $x=1$ (negative to positive) indicates it is not a maximum.
However, at $x = 5$, the function goes from positive ($f(3)=4$) to zero ($f(5)=0$) and back to positive ($f(7)=12$).
This means $f(x)$ touches the x-axis at $(5, 0)$ without crossing it, staying $\ge 0$ in that interval.
Therefore, $(5, 0)$ must be a local minimum.

Question

For the polynomial function $g$, the rate of change of $g$ is increasing for $x < 2$ and decreasing for $x > 2$. Which of the following must be true?

(A) The graph of $g$ has a minimum at $x = 2$.

(B) The graph of $g$ has a maximum at $x = 2$.

(D) The graph of $g$ has a point of inflection at $x = 2$, is concave up for $x < 2$, and is concave down for $x > 2$.

▶️ Answer/Explanation

Detailed solution

The “rate of change” of $g$ refers to the first derivative, $g'(x)$.
If $g'(x)$ is increasing for $x < 2$, then the second derivative $g”(x) > 0$.
A positive second derivative ($g”(x) > 0$) means the graph is concave up.
If $g'(x)$ is decreasing for $x > 2$, then the second derivative $g”(x) < 0$.
A negative second derivative ($g”(x) < 0$) means the graph is concave down.
Since the concavity changes at $x = 2$, there is a point of inflection at $x = 2$.
Therefore, the correct choice is (D).

Question

For $0 \le t \le 16$, 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 $0 \le t \le 16$?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The function $N(t)$ represents the total number of customers, so $N(t) = \int_{0}^{t} R(x) dx$.
Since $R(t)$ is a rate of arrival, $R(t) \ge 0$, which means $N(t)$ must be a non-decreasing function.
The derivative of the total accumulation is the rate, so $N'(t) = R(t)$ and $N”(t) = R'(t)$.
When $R(t)$ is increasing ($0 < t < 4$ and $8 < t < 12$), $N(t)$ is concave up ($N”(t) > 0$).
When $R(t)$ is decreasing ($4 < t < 8$ and $12 < t < 16$), $N(t)$ is concave down ($N”(t) < 0$).
Graph (D) starts at $(0,0)$, always increases, and switches concavity at $t = 4, 8, 12$.
Therefore, the correct graph is (D).

Question

The figure shows the graph of a function $ f $. The zero and extrema are labeled, and the point of inflection is labeled at $ C $. Let $ A, B, C, D, $ and $ E $ represent the $ x $-coordinates of those points. 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 $

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

▶️ Answer/Explanation

Correct Answer:

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

Detailed solution

1. A function is increasing where $ f'(x) > 0 $, meaning the graph rises from left to right.
2. From $ A $ to $ B $, the graph rises to a local maximum at $ B $, so $ f'(x) > 0 $.
3. The point $ C $ is labeled as the point of inflection, so concavity changes at $ C $.
4. Therefore, for all $ x < C $, the graph is concave down ($ f”(x) < 0 $).
5. Since the interval $ A $ to $ B $ lies entirely before $ C $, it is concave down.
6. Hence, $ f $ is increasing and concave down on $ (A, B) $.

Question

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

(A) $0 < t < 4$

(B) $5 < t < 8$

(D) $0 < t < 3$ and $8 < t < 10$

▶️ Answer/Explanation

Answer:

(D) $0 < t < 3$ and $8 < t < 10$

Explanation:

Velocity is decreasing where the graph of $v(t)$ has a negative slope. From $t=0$ to $t=3$, the graph slopes downward continuously. From $t=3$ to $t=5$, the slope is positive (velocity increasing). From $t=5$ to $t=8$, the slope is zero (constant velocity). From $t=8$ to $t=10$, the slope is negative again. Therefore, velocity decreases only on $0 < t < 3$ and $8 < t < 10$.

Detailed solution

1. Velocity decreases when the slope of $v(t)$ is negative.
2. From $0 < t < 3$, the line segment slopes downward ⇒ decreasing.
3. From $3 < t < 5$, the slope is positive ⇒ increasing.
4. From $5 < t < 8$, the slope is $0$ ⇒ constant velocity.
5. From $8 < t < 10$, the slope is negative ⇒ decreasing.
6. Hence, the correct intervals are $0 < t < 3$ and $8 < t < 10$.

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 \le t \le 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

Answer:

(A) $ (3,6) $ only

Detailed solution

1. “Increasing at a decreasing rate” means $ W'(t) > 0 $ and $ W”(t) < 0 $.
2. Thus the graph must be increasing and concave down.
3. From $ t = 0 $ to $ t = 6 $, the function increases.
4. However, from $ (0,3) $, the slope increases (concave up).
5. From $ (3,6) $, the slope decreases to $ 0 $ (concave down).
6. From $ (18,30) $, the function increases but is concave up.
7. Therefore, the only interval satisfying both conditions is $ (3,6) $.

Question

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

(A) $ A $

(B) $ B $

(D) $ D $

▶️ Answer/Explanation

Answer:

(C)

Detailed solution

1. The rate of change of $ g $ at a point is the slope $ g'(x) $.
2. At $ A $, a relative maximum, $ g'(A) = 0 $.
3. At $ D $, near a minimum, the slope is approximately $ 0 $ (slightly positive).
4. At $ B $, the function is decreasing, so $ g'(B) < 0 $, but not steepest.
5. At $ C $, the only point of inflection, the graph changes concavity.
6. At an inflection point on a decreasing curve, the slope is most negative.
7. Therefore, $ g'(C) $ is the smallest (least) value.
8. Hence, the rate of change is least at $ C $.

Question

The figure shows the graph of function $ g $ for $ 0 \le x \le 13 $. 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 \le 13 $. 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

(D) $ [t_C, t_D] $ and $ [t_D, t_E] $

▶️ Answer/Explanation

Answer:

(D) $ [t_C, t_D] $ and $ [t_D, t_E] $

Detailed solution

1. The rate of change of $ g $ is $ g'(x) $. It is decreasing where $ g”(x) < 0 $.
2. $ g”(x) < 0 $ corresponds to the graph being concave down.
3. The graph has exactly one inflection point at $ C $, so concavity changes only at $ t_C $.
4. From $ t_A $ to $ t_C $, the graph is concave up (slope increasing).
5. From $ t_C $ to $ t_E $, the graph is concave down (slope decreasing).
6. Therefore, the rate of change of $ g $ is decreasing on $ [t_C, t_D] $ and $ [t_D, t_E] $.
7. Hence, the correct answer is (D).

Question

The polynomial function $f$ is an odd function with domain $-4 \leq x \leq 4$. The table gives information about values of $f(x)$ and the behavior of the function. What is the absolute maximum value of $f$ on its domain?

(A) $4$

(B) $3$

(D) $0$

▶️ Answer/Explanation

Detailed solution

The correct answer is (B) 3.

From the table, we identify the extrema for $x \ge 0$. The function decreases to a minimum value of $f(2) = -3$ and then increases to a value of $f(4) = 2$.
Since $f$ is an odd function, it satisfies the property $f(-x) = -f(x)$. This means the graph is symmetric with respect to the origin.
A local minimum on the positive side corresponds to a local maximum on the negative side. Therefore, the minimum at $x = 2$ implies a maximum at $x = -2$ given by $f(-2) = -f(2) = -(-3) = 3$.
We must also check the endpoints. We know $f(4) = 2$, which implies $f(-4) = -f(4) = -2$.
Comparing the highest points on the entire domain $[-4, 4]$: the maximum value at $x=4$ is $2$, and the maximum value at $x=-2$ is $3$.
Thus, the absolute maximum value of $f$ is $3$.

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

(D) November

▶️ Answer/Explanation

Answer:

(B) May

Explanation:

The rate of change corresponds to the slope of the temperature curve. The greatest rate of change occurs where the graph is increasing most steeply. From the graph, the temperature rises most rapidly during the spring months. Around May, the curve has its steepest positive slope. In February the slope is small and slightly negative, in August the slope is approximately $0$ (near maximum), and in November the slope is negative. Therefore, the greatest rate of change occurs in May.

Detailed solution

1. The rate of change equals the derivative, represented by the slope of the tangent line.
2. We look for the point where the curve is steepest upward (maximum positive slope).
3. In February, the graph is near a minimum, so slope $\approx 0$.
4. In May, the graph is increasing rapidly, giving the largest positive slope.
5. In August, the graph is near a maximum, so slope $\approx 0$.
6. In November, the graph is decreasing, so slope is negative.
7. Hence, the greatest rate of change occurs in May.

Question

The function $f$ is not explicitly given. The function $g$ is given by $g(x) = f(x+1) – f(x)$. The function $h$ is given by $h(x) = g(x+1) – g(x)$. If $h(x) = -6$ for all values of $x$, which of the following statements must be true?

(A) Because $h$ is negative and constant, the graphs of $g$ and $f$ always have negative slope.

(B) Because $h$ is negative and constant, the graphs of $g$ and $f$ are concave down.

(D) Because $h$ is negative and constant, $g$ is decreasing, and the graph of $f$ is concave down.

▶️ Answer/Explanation

Detailed solution

We analyze the definitions of $g$ and $h$ to determine the properties of $f$.
1. We are given $h(x) = g(x+1) – g(x) = -6$. Since the difference is negative, $g(x+1) < g(x)$, which means $g$ is strictly decreasing.
2. The function $g(x)$ represents the first difference (approximate slope) of $f(x)$, and $h(x)$ represents the second difference of $f(x)$.
3. In calculus, a constant negative second difference (like $h(x) = -6$) indicates that the function $f$ is concave down (like an inverted parabola, $y = -3x^2$).
4. While $f$ is concave down, it does not necessarily always have a negative slope (e.g., the left side of an inverted parabola has a positive slope).
5. Therefore, the only statements that must be true are that $g$ is decreasing and $f$ is concave down.
Conclusion: This corresponds to option (D).

Question

The function $ f $ has a negative average rate of change on every interval of $ x $ in the interval $ 0 \le x \le 10 $. The function $ g $ has a negative average rate of change on every interval of $ x $ in the interval $ 0 \le x < 5 $, and a positive average rate of change on every interval of $ x $ in the interval $ 5 < x \le 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 \le x \le 10 $?

(A) $ h $ is decreasing on $ 0 \le x \le 10 $.

(B) $ h $ is decreasing on $ 0 \le x < 5 $; $ h $ is increasing on $ 5 < x \le 10 $.

(D) $ h $ is decreasing on $ 0 \le x < 5 $; $ h $ can be increasing, decreasing, or both increasing and decreasing on $ 5 < x \le 10 $.

▶️ Answer/Explanation

Detailed solution

Step 1: The condition “negative average rate of change on every interval” implies the function is strictly decreasing. Thus, $ f(x) $ is decreasing on $ 0 \le x \le 10 $.

Step 2: Similarly, $ g(x) $ is decreasing on $ 0 \le x < 5 $ and increasing (positive average rate of change) on $ 5 < x \le 10 $.

Step 3: On the interval $ 0 \le x < 5 $, $ h(x) = f(x) + g(x) $ is the sum of two decreasing functions, meaning $ h(x) $ must be decreasing.

Step 4: On the interval $ 5 < x \le 10 $, $ f(x) $ is decreasing while $ g(x) $ is increasing. The direction of $ h(x) $ depends on which function changes more rapidly.

Step 5: Since the magnitudes of the rates of change are unknown, $ h(x) $ is indeterminate on this interval; it could increase, decrease, or fluctuate.

Step 6: Therefore, statement (D) correctly identifies that while the first part is definitely decreasing, the second part can be any of the listed behaviors.

Correct Option: (D)

Question

A second hand on a clock rotates clockwise at a constant speed around the clock. However, Charlie bought a new clock that is dysfunctional, where the second hand completes a cycle every (30) seconds. The clock is (24) inches in diameter and is (5) feet above the ground. The distance between the point at which the second hand is pointing to and the ground is periodically increasing and decreasing. The second hand starts at (t = 0) pointing straight up towards the number (12).

(A) This situation can be represented with a sinusoidal function in the form $h(t) = a \cos( b(t + c) ) + d$, where $t$ is the number of seconds after starting the clock and $h(t)$ is the height of the end of the second hand in inches. Find the function.

(B) When does the arrow of the second hand become $72 – 6\sqrt{2}$ inches above the ground on the interval $0 < t \leq 30$?

(C) i. Between $0 \leq t \leq 30$, does the graph of $h(t)$ have a point of inflection? If so, identify all the points of inflection between the time intervals. If it does not, explain why using your knowledge of a cosine function and its continuity.

ii. Explain the meaning of these points in context of the problem using the definition of the rate of change and the rate the rate is changing.

Most-appropriate topic codes (CED):

• TOPIC 3.7: Sinusoidal Function Context and Data Modeling — part (A)
• TOPIC 3.10: Trigonometric Equations and Inequalities — part (B)
• TOPIC 1.1: Change in Tandem — part (C)

▶️ Answer/Explanation

Detailed solution

(A)
First, determine the physical parameters of the clock to find the sinusoidal constants.
The diameter is (24) inches, so the radius is (12) inches. This corresponds to the amplitude, (a = 12).
The center of the clock is (5) feet above the ground, which converts to (60) inches. This is the midline, (d = 60).
The second hand completes a cycle every (30) seconds. The period is (30), so the frequency coefficient is (b = \dfrac{2\pi}{30} = \dfrac{\pi}{15}).
At (t = 0), the hand is pointing straight up (maximum height). Since the cosine function starts at a maximum, there is no phase shift, so (c = 0).
Thus, the function is:
(h(t) = 12 \cos\left( \dfrac{\pi}{15} t \right) + 60)

(B)
Set $h(t)$ equal to the given height and solve for $t$:
$12 \cos\left( \dfrac{\pi}{15} t \right) + 60 = 72 – 6\sqrt{2}$
$12 \cos\left( \dfrac{\pi}{15} t \right) = 12 – 6\sqrt{2}$
$\cos\left( \dfrac{\pi}{15} t \right) = 1 – \dfrac{\sqrt{2}}{2} = \dfrac{2 – \sqrt{2}}{2}$
Let $\theta = \dfrac{\pi}{15} t$. We need to solve $\cos \theta = \dfrac{2 – \sqrt{2}}{2}$.
$\theta = \pm \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right) + 2\pi k$.
For the interval $0 < t \leq 30$, we look for solutions in $(0, 2\pi]$.
The two solutions for $\theta$ are $\theta_1 = \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right)$ and $\theta_2 = 2\pi – \theta_1$.
Converting back to time $t = \dfrac{15}{\pi} \theta$:
$t_1 = \dfrac{15}{\pi} \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right)$ and $t_2 = 30 – t_1$.

(C) i.
Yes, the graph of $h(t)$ has points of inflection. The function $h(t)$ is a smooth, continuous cosine wave.
Points of inflection on a sinusoidal graph occur where the graph crosses its midline (concavity changes from up to down or vice versa). This happens when:
$\cos\left( \dfrac{\pi}{15} t \right) = 0$
$\dfrac{\pi}{15} t = \dfrac{\pi}{2} + k\pi$
$t = 15\left( \dfrac{1}{2} + k \right) = 7.5 + 15k$
For the interval $0 \leq t \leq 30$:
If $k=0$, $t = 7.5$.
If $k=1$, $t = 22.5$.
The points of inflection are at $t = 7.5$ seconds and $t = 22.5$ seconds.

(C) ii.
In this context:
The rate of change represents the vertical velocity of the tip of the second hand.
The rate the rate is changing represents the vertical acceleration.
A point of inflection occurs where the concavity changes, which means the vertical acceleration is transitioning from positive to negative (or vice versa) and is instantaneously zero. Physically, these are the moments when the tip of the second hand is at the same height as the center of the clock (midline). At these specific points, the vertical speed (rate of change) is at its absolute maximum magnitude.

Question

A market analyst working for a small appliance manufacturer finds that if the firm produces and sells $x$ blenders annually, the total profit (in dollars) is $P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888$. (4 marks each part)

a. Use a graphing device to help graph the polynomial function $P$.

b. Find the average rate of change of $P$ between two relative (local) extrema when $0 < x < 200$.

c. Find the equation of the secant line of the graph of $P$ between the two points in part b.

d. The inflection point of a function is the point on the graph where the graph changes concavity. Find out the inflection point of the graph of $P$.

e. How will the rate of change vary before and after the inflection point?

▶️ Answer/Explanation

Detailed solution

a. Graphing the function
Using a graphing utility for $P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888$ on the interval $[0, 200]$ reveals a cubic curve shape.
The graph starts with a slight dip to a local minimum near $x=0$, then rises steeply to a local maximum near $x=180$, before falling again.

b. Average rate of change between extrema
First, find the derivative: $P'(x) = -0.0039x^2 + 0.7014x – 0.4591$.
Set $P'(x) = 0$ and use the quadratic formula to find the extrema: $x \approx 0.66$ (local min) and $x \approx 179.19$ (local max).
Calculate the profit at these points: $P(0.66) \approx -422.04$ and $P(179.19) \approx 3276.57$.
The average rate of change is: $\frac{3276.57 – (-422.04)}{179.19 – 0.66} = \frac{3698.61}{178.53} \approx 20.72$.

c. Equation of the secant line
The slope $m$ was found in part (b) to be approximately $20.72$.
Using the point-slope form $y – y_1 = m(x – x_1)$ with the minimum point $(0.66, -422.04)$:
$y – (-422.04) = 20.72(x – 0.66)$
$y = 20.72x – 13.68 – 422.04$
The equation is approximately $y = 20.72x – 435.72$.

d. Inflection point
Find the second derivative: $P”(x) = -0.0078x + 0.7014$.
Set $P”(x) = 0$ to find the change in concavity: $0 = -0.0078x + 0.7014 \Rightarrow x = \frac{0.7014}{0.0078} \approx 89.92$.
Find the corresponding y-value: $P(89.92) \approx 1427.27$.
The inflection point is approximately $(89.92, 1427.27)$.

e. Variation of rate of change
The rate of change is represented by the derivative, $P'(x)$.
Before the inflection point ($x < 89.92$), the graph is concave up ($P”(x) > 0$), so the rate of change is increasing.
After the inflection point ($x > 89.92$), the graph is concave down ($P”(x) < 0$), so the rate of change is decreasing.

\(x\)	\(g(x)\)
1	1
2	2
3	4
4	8
5	16

\(x\)	\(g(x)\)
1	-20
2	-10
3	-5
4	5
5	10

\(x\)	\(g(x)\)
1	1
2	5
3	7
4	5
5	1

\(x\)	\(g(x)\)
1	7
2	3
3	1
4	3
5	7