AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

For a polynomial function $ f $, $ \lim_{x \to -\infty} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = -\infty $. Which of the following must be true about $ f $?

(A) The degree of $ f $ is even, and the leading coefficient is negative.
(B) The degree of $ f $ is even, and the leading coefficient is positive.
(C) The degree of $ f $ is odd, and the leading coefficient is negative.
(D) The degree of $ f $ is odd, and the leading coefficient is positive.

▶️ Answer/Explanation

Detailed solution

End behavior: as $ x \to -\infty $, $ f(x) \to \infty $ (up to the left); as $ x \to \infty $, $ f(x) \to -\infty $ (down to the right).
For odd degree polynomials: ends go in opposite directions.
Leading coefficient negative ⇒ up-left, down-right. Leading coefficient positive ⇒ down-left, up-right.
Given up-left, down-right ⇒ odd degree, negative leading coefficient.
✅ Answer: (C)

Question

The polynomial function $ p $ is given by $ p(x) = -4x^5 + 3x^2 + 1 $. Which of the following statements about the end behavior of $ p $ is true?

(A) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is even; therefore, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = \infty $.
(B) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
(C) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = -\infty $ and $ \lim_{x \to \infty} p(x) = \infty $.
(D) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = -\infty $ and $ \lim_{x \to \infty} p(x) = \infty $.

▶️ Answer/Explanation

Detailed solution

Leading term: $ -4x^5 $ → negative coefficient, degree 5 (odd).
For odd degree, negative leading coefficient: up-left, down-right.
So $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
This matches option (B).
✅ Answer: (B)

Question

The polynomial function $ f $ is given by $ f(x) = (x – 4)(3x – 1)^2 $. Which of the following descriptions of $ f $ is true?

(A) $ f $ is a polynomial of degree 2 with a leading coefficient of 3.
(B) $ f $ is a polynomial of degree 2 with a leading coefficient of 9.
(C) $ f $ is a polynomial of degree 3 with a leading coefficient of 3.
(D) $ f $ is a polynomial of degree 3 with a leading coefficient of 9.

▶️ Answer/Explanation

Detailed solution

Factor $ (3x – 1)^2 $ has degree 2, factor $ (x – 4) $ has degree 1 ⇒ total degree = 3.
Leading term from expanding: $ (3x)^2 \cdot x = 9x^3 $ ⇒ leading coefficient 9.
✅ Answer: (D)

Question

A polynomial function has the form $ p(x) = ax^j + bx^k $, where $ a $ and $ b $ are nonzero constants, $ j $ and $ k $ are nonnegative integers. Which of the following conditions guarantees that $ p $ is an even function?

(A) $ k = 0 $
(B) $ j = 2k $ and $ k $ is even.
(C) $ j + k $ is even.
(D) $ j \cdot k $ is even.

▶️ Answer/Explanation

Detailed solution

Even function means $ p(-x) = p(x) $.
This requires $ j $ and $ k $ both even, because $ (-x)^n = x^n $ only if $ n $ is even.
Check options:
(A) $ k=0 $ (0 is even) but $ j $ could be odd ⇒ not guaranteed even.
(B) $ j = 2k $ and $ k $ even ⇒ $ j $ even (since even × integer = even). Both exponents even ⇒ $ p $ even.
(C) $ j+k $ even: e.g., $ j=1, k=3 $ ⇒ both odd ⇒ $ p $ not even.
(D) $ j \cdot k $ even: e.g., $ j=1, k=2 $ ⇒ $ j $ odd ⇒ not even.
Thus only (B) guarantees both exponents even.
✅ Answer: (B)

Question

The polynomial function $ P $ is given by $ P(x) = 3x(x + 1)^2(x – a) $, where $ a $ is a real number. Which of the following could be the graph of $ y = P(x) $?

▶️ Answer/Explanation

Detailed solution

Given $ P(x) = 3x(x + 1)^2(x – a) $.
For both $ x = -1 $ and $ x = 0 $ to have even multiplicity (graph tangent to x-axis at both), $ x = 0 $ must be multiplicity 2 ⇒ $ a = 0 $ makes $ P(x) = 3x^2(x + 1)^2 $, so both zeros multiplicity 2.
The correct graph must show tangency at $ x = 0 $ and $ x = -1 $, no crossing there, and degree 4, positive leading coefficient 3 ⇒ ends up.
Match to the option with that shape.
✅ Answer: (D)

Question

The polynomial function $ f $ is given by $ f(x) = ax^4 + bx^3 + cx^2 + dx + k $, where $ a \neq 0 $ and $ b, c, d, $ and $ k $ are constants. Which of the following statements about $ f $ is true?

(A) $ f $ has both a global maximum and a global minimum.
(B) $ f $ has either a global maximum or a global minimum, but not both.
(C) $ f $ has neither a global maximum nor a global minimum.
(D) The nature of a global maximum or a global minimum for $ f $ cannot be determined without more information about $ b, c, d, $ and $ k $.

▶️ Answer/Explanation

Detailed solution

$ f $ is a quartic polynomial (degree 4, even).
For even-degree polynomials with leading coefficient $ a \neq 0 $:
– If $ a > 0 $, as $ x \to \pm\infty $, $ f(x) \to +\infty $ ⇒ global minimum exists, no global maximum.
– If $ a < 0 $, as $ x \to \pm\infty $, $ f(x) \to -\infty $ ⇒ global maximum exists, no global minimum.
Thus $ f $ has either a global max or a global min, but not both.
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = 5x^6 – 2x^3 – 3 $. Which of the following describes the end behavior of $ f $?

(A) $ \displaystyle \lim_{x \to -\infty} f(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $
(B) $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = \infty $
(C) $ \displaystyle \lim_{x \to -\infty} f(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = \infty $
(D) $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial is determined by its leading term when the degree is even or odd and the leading coefficient is positive or negative.
For $ f(x) = 5x^6 – 2x^3 – 3 $:
• Degree = 6 (even).
• Leading coefficient = 5 (positive).

For even degree and positive leading coefficient:
As $ x \to \pm\infty $, $ 5x^6 \to +\infty $.
Thus:
\[ \lim_{x \to -\infty} f(x) = \infty \] \[ \lim_{x \to \infty} f(x) = \infty \]
This matches option (B).
✅ Answer: (B)

Question

In the odd function $f(x)$, the $\lim_{x \to 3^-} f(x) = -\infty$. Which one of the following must also be true?

a. $\lim_{x \to -3^-} f(x) = -\infty$

b. $\lim_{x \to -3^-} f(x) = \infty$

c. $\lim_{x \to -3^+} f(x) = -\infty$

d. $\lim_{x \to -3^+} f(x) = \infty$

▶️ Answer/Explanation

Detailed solution

An odd function satisfies the condition $f(-x) = -f(x)$, which implies symmetry about the origin.
The given limit $\lim_{x \to 3^-} f(x) = -\infty$ describes the behavior as $x$ approaches $3$ from the left.
To find the behavior near $-3$, we substitute $x$ with $-t$, so as $x \to -3^+$, then $t \to 3^-$.
Using the odd property: $f(x) = f(-t) = -f(t)$.
Therefore, $\lim_{x \to -3^+} f(x) = \lim_{t \to 3^-} [-f(t)] = -(-\infty) = \infty$.
This matches option d.
Correct Option: d

Question

The function is a polynomial with the following end behavior:

$\lim_{x\to-\infty} g(x) = -\infty \quad \text{and} \quad \lim_{x\to\infty} g(x) = -\infty$

Which of the following could be an expression for $g(x)$?

(A) $-3x^2 + 4x^2 + 1$

(B) $-x^4 + 2x^3 + 6$

(D) $4x^6 + x^3 – 5$

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial function is determined by its leading term (the term with the highest degree).
Since $\lim_{x\to\infty} g(x) = -\infty$ and $\lim_{x\to-\infty} g(x) = -\infty$, the graph points downwards on both ends.
This behavior indicates that the polynomial must have an even degree and a negative leading coefficient.
(A) simplifies to $x^2 + 1$: Even degree, positive coefficient (Ends $\to \infty$). Incorrect.
(B) $-x^4 + 2x^3 + 6$: Even degree (4), negative coefficient (-1). Ends $\to -\infty$. Correct.
(C) $2x^3 – 7x^2 + 4x$: Odd degree. Ends go in opposite directions. Incorrect.
(D) $4x^6 + x^3 – 5$: Even degree, positive coefficient (Ends $\to \infty$). Incorrect.

Question

The polynomial function $p$ is given by $p(x) = ax^n + x – 4$ where $a$ and $n$ are constants and $n > 1$. If the end behavior of $p$ is given by the statements $\lim_{x \to -\infty} p(x) = \infty$ and $\lim_{x \to \infty} p(x) = -\infty$, which of the following could be the values of $a$ and $n$?

(A) $a = -3$ and $n = 5$

(B) $a = -3$ and $n = 4$

(D) $a = 3$ and $n = 4$

▶️ Answer/Explanation

Detailed solution

The end behavior of the polynomial $p(x)$ is determined by its leading term, $ax^n$.
Since $\lim_{x \to \infty} p(x) = -\infty$, the leading term must be negative as $x$ gets large, which implies $a < 0$.
This condition eliminates options (C) and (D) because $a$ must be negative.
Since $\lim_{x \to -\infty} p(x) = \infty$, the function rises to the left. With $a < 0$, $x^n$ must be negative.
For $x^n$ to be negative when $x$ is negative, the degree $n$ must be an odd integer.
Looking at the remaining options, only (A) has an odd value for $n$ ($n=5$).
Therefore, the correct choice is $a = -3$ and $n = 5$.

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) $3$
(C) $2$
(D) $0$

▶️ Answer/Explanation

Detailed solution

Since $f$ is an odd function, it satisfies the property $f(-x) = -f(x)$ for all $x$ in the domain.
From the table, the minimum value on the positive domain $[0, 4]$ is $f(2) = -3$.
By the odd property, $f(-2) = -f(2) = -(-3) = 3$.
The value $f(4) = 2$ corresponds to $f(-4) = -f(4) = -2$.
In the interval $2 < x < 3$, $f(x)$ increases from $-3$ to $0$, so on $-3 < x < -2$, $f(x)$ decreases from $0$ to $3$.
The maximum value in the positive interval is $f(4) = 2$, while the odd property reveals a higher value of $3$ at $x = -2$.
Comparing all critical points and boundaries, the absolute maximum value is $3$.
Therefore, the correct option is (B).

Question

The polynomial function $p$ is given by $p(x) = ax^n + b$, where $a$ and $b$ are nonzero constants and $n$ is a positive integer. As the input values of $p$ increase without bound, the output values of $p$ decrease without bound. Which of the following must be true?

A. $a$ is negative because $\lim_{x \to \infty} p(x) = -\infty$.

B. $a$ is positive because $\lim_{x \to \infty} p(x) = -\infty$.

C. $a$ is negative because $\lim_{x \to \infty} p(x) = \infty$.

D. $a$ is positive because $\lim_{x \to \infty} p(x) = \infty$.

▶️ Answer/Explanation

Detailed solution

The phrase “input values of $p$ increase without bound” corresponds to the limit as $x \to \infty$.
The phrase “output values of $p$ decrease without bound” corresponds to the limit equaling $-\infty$.
Therefore, the condition is defined as $\lim_{x \to \infty} p(x) = -\infty$.
Considering the leading term, $\lim_{x \to \infty} (ax^n + b) = \lim_{x \to \infty} ax^n$.
Since $n$ is a positive integer, $x^n \to \infty$ as $x$ increases.
For the product $ax^n$ to approach $-\infty$, the coefficient $a$ must be negative.
Thus, the correct statement is that $a$ is negative because $\lim_{x \to \infty} p(x) = -\infty$.

Question

Let $ k $ be a polynomial function with the following end behavior: \[ \lim_{x \to -\infty} k(x) = \infty \text{ and } \lim_{x \to \infty} k(x) = -\infty \] Which of the following could be an expression for $ k(x) $?

(A) $ 4x^5 + x^2 + 3 $

(B) $ 3x^4 – x^3 + 5 $

(D) $ -5x^6 + x^3 + 6 $

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial function is determined by its leading term (the term with the highest degree).
The problem states that as $ x \to -\infty $, $ k(x) \to \infty $ (starts up) and as $ x \to \infty $, $ k(x) \to -\infty $ (ends down).
Since the ends point in opposite directions, the degree of the polynomial must be odd.
Since the function goes to $ -\infty $ as $ x $ approaches positive infinity (right side goes down), the leading coefficient must be negative.
Looking at the options, we analyze the leading terms:
(A) $ 4x^5 $: Odd degree, positive coefficient (Ends: Down/Up). Incorrect.
(B) $ 3x^4 $: Even degree, positive coefficient (Ends: Up/Up). Incorrect.
(C) $ -2x^3 $: Odd degree, negative coefficient (Ends: Up/Down). Correct.
(D) $ -5x^6 $: Even degree, negative coefficient (Ends: Down/Down). Incorrect.

Question

For a polynomial function $ f $, $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $. Which of the following must be true about $ f $?

(A) The degree of $ f $ is even, and the leading coefficient is negative.

(B) The degree of $ f $ is even, and the leading coefficient is positive.

(D) The degree of $ f $ is odd, and the leading coefficient is positive.

▶️ Answer/Explanation

Answer:

(C) The degree of $ f $ is odd, and the leading coefficient is negative.

Explanation:

The end behavior of a polynomial is determined by its leading term $ a x^n $. If the limits at $ -\infty $ and $ \infty $ are opposite in sign, the degree must be odd. Since $ f(x) \to -\infty $ as $ x \to \infty $, the leading coefficient must be negative.

Detailed solution

Line 1: The end behavior of $ f(x) $ depends on its leading term $ a x^n $.
Line 2: If $ n $ is even, both ends of the graph go in the same direction.
Line 3: Here, $ \lim_{x \to -\infty} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = -\infty $, so the ends differ.
Line 4: Therefore, the degree $ n $ must be odd.
Line 5: For an odd degree polynomial, if $ a < 0 $, then $ f(x) \to -\infty $ as $ x \to \infty $.
Line 6: This matches the given condition, so the leading coefficient is negative.
Line 7: Hence, the correct answer is (C).

Question

For the polynomial function $ g $, $ \displaystyle \lim_{x \to -\infty} g(x) = -\infty $. Which of the following expressions could define $ g(x) $?

(A) $ -5x – 2x^4 $

(B) $ -5x^2 – 2x^7 $

(D) $ -x^3 + 2x^4 $

▶️ Answer/Explanation

Answer:

(A)

Detailed solution

1. The end behavior of a polynomial depends on its highest-degree term.
2. In (A), the leading term is $ -2x^4 $.
3. Since $ x^4 $ is even, $ x^4 \to +\infty $ as $ x \to -\infty $.
4. Multiplying by $ -2 $ gives $ -2x^4 \to -\infty $.
5. In (B) and (C), the leading terms are odd powers with negative coefficients, giving $ +\infty $ as $ x \to -\infty $.
6. In (D), the leading term is $ 2x^4 $, which gives $ +\infty $.
7. Therefore, only option (A) satisfies $ \lim_{x \to -\infty} g(x) = -\infty $.

Question

The function $ f $ is given by $ f(x) = -2x^{7} + 5x^{4} + 6x^{2} – 3$. Which of the following correctly describes the end behavior of $ f $ as the input values increase without bound?

(A) $\displaystyle \lim_{x \to \infty} f(x) = \infty$

(B) $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

(D) $\displaystyle \lim_{x \to -\infty} f(x) = -\infty$

▶️ Answer/Explanation

Answer:

(B) $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

Detailed solution

1. The highest-degree term is $ -2x^{7} $.
2. End behavior of a polynomial is determined by its leading term.
3. As $ x \to \infty $, $ x^{7} \to \infty $.
4. Therefore, $ -2x^{7} \to -\infty $.
5. Lower-degree terms $ 5x^{4}, 6x^{2}, -3 $ become insignificant compared to $ x^{7} $.
6. Hence, $ f(x) \to -\infty $ as $ x \to \infty $.
7. Therefore, the correct answer is option (B).

Question

The function $ f $ is given by $ f(x) = 5x^6 – 2x^3 – 3 $. Which of the following describes the end behavior of $ f $?

(A) $\displaystyle \lim_{x \to -\infty} f(x) = -\infty$ and $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

(B) $\displaystyle \lim_{x \to -\infty} f(x) = \infty$ and $\displaystyle \lim_{x \to \infty} f(x) = \infty$

(D) $\displaystyle \lim_{x \to -\infty} f(x) = \infty$ and $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

▶️ Answer/Explanation

Answer:

(B)

Explanation:

The leading term of the polynomial is $5x^6$. Since the degree is even ($6$) and the leading coefficient $5$ is positive, the function rises to positive infinity on both ends. Therefore, \[ \lim_{x \to -\infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to \infty} f(x) = \infty. \]

Detailed solution

1. The highest-degree term is $5x^6$.
2. End behavior depends only on the leading term.
3. The degree $6$ is even.
4. The leading coefficient $5 > 0$.
5. For even degree with positive coefficient, both ends rise.
6. Thus, $f(x) \to \infty$ as $x \to \pm\infty$.
7. Therefore, the correct answer is (B).

Question

The polynomial function is given by $ p(x) = -4x^5 + 3x^2 + 1 $. Which of the following statements about the end behavior of $ p $ is true?

(A) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is even; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

(B) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = -\infty $

(C) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

(D) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

▶️ Answer/Explanation

Correct Answer: (B)

The leading term of the polynomial is $ -4x^5 $. Since the coefficient $ -4 $ is negative and the degree $ 5 $ is odd, the graph rises to $ \infty $ as $ x \to -\infty $ and falls to $ -\infty $ as $ x \to \infty $. Therefore, option (B) correctly describes the end behavior.

Detailed solution

1. The given polynomial is $ p(x) = -4x^5 + 3x^2 + 1 $.
2. The leading term is $ -4x^5 $.
3. The coefficient $ -4 $ is negative.
4. The degree $ 5 $ is odd.
5. For an odd degree with negative leading coefficient: left end rises, right end falls.
6. Thus, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
7. Hence, the correct statement is (B).

Question

The polynomial function $ f $ is given by $ f(x) = (x – 4)(3x – 1)^2$. Which of the following descriptions of $ f $ is true?

(A) $ f $ is a polynomial of degree $ 2 $ with a leading coefficient of $ 3 $.

(B) $ f $ is a polynomial of degree $ 2 $ with a leading coefficient of $ 9 $.

(D) $ f $ is a polynomial of degree $ 3 $ with a leading coefficient of $ 9 $.

▶️ Answer/Explanation

Answer:

(D) $ f $ is a polynomial of degree $ 3 $ with a leading coefficient of $ 9 $.

Explanation:

The factor $ (x – 4) $ has degree $ 1 $, and $ (3x – 1)^2 $ has degree $ 2 $. Thus, the total degree is $ 1 + 2 = 3 $. The leading term of $ (3x – 1)^2 $ is $ 9x^2 $. Multiplying by $ x $ from $ (x – 4) $ gives the leading term $ 9x^3 $. Therefore, the leading coefficient is $ 9 $.

Detailed solution

1. Given $ f(x) = (x – 4)(3x – 1)^2 $.
2. Degree of $ (x – 4) = 1 $.
3. Degree of $ (3x – 1)^2 = 2 $.
4. Total degree $ = 1 + 2 = 3 $.
5. Leading term of $ (3x – 1)^2 = 9x^2 $.
6. Multiply by leading term $ x $: $ x \cdot 9x^2 = 9x^3 $.
7. Hence, leading coefficient $ = 9 $.
8. Correct option is (D).

Question

A polynomial function has the form $ p(x) = ax^j + bx^k $, where $a$ and $b$ are nonzero constants, and $j$ and $k$ are nonnegative integers. Which of the following conditions guarantees that $p$ is an even function?

(A) $ k = 0 $

(B) $ j = 2k $ and $ k $ is even

(D) $ j \cdot k $ is even

▶️ Answer/Explanation

Answer:

(B)

Explanation:

A function is even if $ p(-x) = p(x) $. For $ p(x) = ax^j + bx^k $, we have \[ p(-x) = a(-x)^j + b(-x)^k. \] This equals $ ax^j + bx^k $ only when both exponents $j$ and $k$ are even. Option (B) guarantees this because if $k$ is even and $j = 2k$, then $j$ is also even. Hence both exponents are even, making $p(x)$ an even function.

Detailed solution

1. A function is even if $ p(-x) = p(x) $.
2. Compute $ p(-x) = a(-x)^j + b(-x)^k $.
3. For equality, both $ (-x)^j = x^j $ and $ (-x)^k = x^k $.
4. This occurs only when $ j $ and $ k $ are even integers.
5. In option (B), $ k $ is even and $ j = 2k \Rightarrow j $ is even.
6. Therefore, both exponents are even, guaranteeing an even function.

Question

An increasing, continuous function $ g(x) $ has outputs of some values of $ x $ shown above. $ f(x) $ is a function shown to the right as the graph.

(A) i. Estimate the value of $ g(f^{-1}(1)) $ for all values if it is defined.

ii. If $ h(x) = \frac{g(x)}{f(x)} $, then define any discontinuities, their type, and their location when $ x > 0 $.

(B) Suppose $ j(x) = \frac{1}{4.99 – \ln(x)} \cdot e^{2 \sin (\sqrt{x})} $ could be used to model $ f(x) $.

i. Determine the end behavior of $ j $ when it increases without bound. Express your answer with the mathematical definition of a limit if it exists. If it does not exist, explain why.

ii. Determine the vertical asymptote of $ j $ when it approaches $ 0 $ from the right. Express your answer with the mathematical definition of a limit if it exists. If it does not, explain why.

(C) Determine if $ g(x) $, using the values in the table, can be best modeled with a linear, quadratic, exponential, or logarithmic function. Explain your answer based on the relationship between the change in the output values of $ g $ and the change in the input values of $ g $.

Most-appropriate topic codes (CED):

• TOPIC 2.8: Inverses of Exponential Functions — part (A)i
• TOPIC 1.10: Rational Functions and Holes — part (A)ii
• TOPIC 1.7: Rational Functions and End Behavior — part (B)i
• TOPIC 1.9: Rational Functions and Vertical Asymptotes — part (B)ii
• TOPIC 2.10: Logarithmic Function Context and Data Modeling — part (C)

▶️ Answer/Explanation

Detailed solution

(A)

i. The graph of $ f(x) $ shows that $ f(2) \approx 1 $, so $ f^{-1}(1) \approx 2 $. From the table, $ g(2) = 1 $. Thus, $ g(f^{-1}(1)) \approx 1 $.
Since $ f(x) $ is continuous and increasing, the inverse is defined for values in its range.
The estimate is a single value, $ 1 $, as the graph aligns precisely with the point. No other values exist due to the strictly increasing nature of $ f(x) $.

ii. The graph of $ f(x) $ crosses zero at $ x=1 $, so $ f(1)=0 $, and from the table, $ g(1)=0 $.
Thus, $ h(1) $ is undefined ($ 0/0 $ form).
The limit as $ x \to 1 $ of $ h(x) $ exists and equals $ 1 $, since both $ g(x) $ and $ f(x) $ resemble $ \log_2(x) $, making $ h(x)=1 $ elsewhere.
This indicates a removable discontinuity at $ x=1 $. No other discontinuities for $ x>0 $, as $ f(x) \neq 0 $ elsewhere in the domain.
The location is $ x=1 $, type: removable.

(B)

i. As $ x \to \infty $, $ \ln(x) \to \infty $, so $ 4.99 – \ln(x) \to -\infty $, and $ \frac{1}{4.99 – \ln(x)} \to 0^- $.
The term $ e^{2 \sin(\sqrt{x})} $ oscillates between $ e^{-2} $ and $ e^{2} $, remaining bounded.
By the squeeze theorem, since the amplitude approaches $ 0 $, $ \lim_{x \to \infty} j(x) = 0 $.
The end behavior is that $ j(x) $ approaches $ 0 $. The limit exists.

ii. As $ x \to 0^+ $, $ \ln(x) \to -\infty $, so $ 4.99 – \ln(x) \to +\infty $, and $ \frac{1}{4.99 – \ln(x)} \to 0^+ $.
Also, $ \sqrt{x} \to 0^+ $, $ \sin(\sqrt{x}) \to 0^+ $, so $ e^{2 \sin(\sqrt{x})} \to e^0 = 1 $.
Thus, $ j(x) \to 0 \cdot 1 = 0 $.
The limit is $ \lim_{x \to 0^+} j(x) = 0 $.
No vertical asymptote at $ x=0 $, as the function approaches a finite value, not $ \pm \infty $.

(C)

The table shows $ g(x) $ values: at $ x=0.5, 1, 2, 4, 7, 8 $, $ g(x)=-1, 0, 1, 2, 2.807, 3 $.
When $ x $ doubles ($ 0.5 $ to $ 1 $, $ 1 $ to $ 2 $, $ 2 $ to $ 4 $, $ 4 $ to $ 8 $), $ \Delta g = +1 $ consistently.
This pattern matches logarithmic functions, where outputs increase by a constant for multiplicative input changes.
Linear would require constant $ \Delta g $ for constant $ \Delta x $, but $ \Delta x $ varies while $ \Delta g =1 $ for doublings.
Quadratic would show constant second differences, but here first differences are not constant.
Exponential would show constant ratios in outputs for additive input changes, which does not fit.
Thus, best modeled by a logarithmic function like $ g(x) = \log_2(x) $.

Question

Let $f$ be an increasing function defined for $x > 0$. The table gives values for $f(x)$ at selected values of $x$. The function $g$ is given by $g(x) = 0.25x^3 – 9.5x^2 + 110x – 399$.

Part A

(i) The function $h$ is defined by $h(x) = (g \circ f)(x) = g(f(x))$. Find the value of $h(8)$ as a decimal approximation, or indicate that it is not defined.

(ii) Find the value of $f^{-1}(20)$, or indicate that it is not defined.

Part B

(i) Find all values of $x$, as decimal approximations, for which $g(x) = -45$, or indicate there are no such values.

(ii) Determine the end behavior of $g$ as $x$ increases without bound. Express your answer using the mathematical notation of a limit.

Part C

(i) Use the table of values of $f(x)$ to determine if $f$ is best modeled by a linear, quadratic, exponential, or logarithmic function.

(ii) Give a reason for your answer based on the relationship between the change in the output values of $f$ and the change in the input values of $f$.

▶️ Answer/Explanation

Detailed solution

Part A

(i) From the table, we find $f(8) = 15$.

Substitute $15$ into the function $g(x)$: $h(8) = g(15) = 0.25(15)^3 – 9.5(15)^2 + 110(15) – 399$.

$h(8) = 0.25(3375) – 9.5(225) + 1650 – 399$.

$h(8) = 843.75 – 2137.5 + 1650 – 399$.

$h(8) = -42.75$.

(ii) To find $f^{-1}(20)$, we look for the $x$ value where $f(x) = 20$.

From the table, $f(16) = 20$.

Therefore, $f^{-1}(20) = 16$.

Part B

(i) We solve the equation $0.25x^3 – 9.5x^2 + 110x – 399 = -45$.

Set the equation to zero: $0.25x^3 – 9.5x^2 + 110x – 354 = 0$.

Using numerical methods or a graphing calculator, the real solutions are approximately:

$x \approx 5.242$

$x \approx 12.188$

$x \approx 20.570$

(ii) The end behavior of a polynomial is determined by its leading term, $0.25x^3$.

Since the leading coefficient is positive and the degree is odd, as $x \to \infty$, $g(x) \to \infty$.

The limit notation is $\lim_{x \to \infty} g(x) = \infty$.

Part C

(i) The function $f$ is best modeled by a logarithmic function.

(ii) In a logarithmic model, constant changes in the output values correspond to proportional changes in the input values.

As the output $f(x)$ increases by a constant $5$ ($0, 5, 10, 15 \dots$),

The input $x$ values are multiplied by a constant factor of $2$ ($1, 2, 4, 8 \dots$).

This constant ratio of inputs for constant additions of outputs is the hallmark of logarithmic growth.

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.

Question

A graphing calculator is required for these questions.

Let $f$ be an increasing function defined for $x \ge 0$. The table gives values of $f(x)$ at selected values of $x$. The function $g$ is given by $g(x) = \frac{x^3 – 14x – 27}{x+2}$.

(A) (i) The function $h$ is defined by $h(x) = (g \circ f)(x) = g(f(x))$. Find the value of $h(5)$ as a decimal approximation, or indicate that it is not defined.

(ii) Find the value of $f^{-1}(4)$, or indicate that it is not defined.

(B) (i) Find all values of $x$, as decimal approximations, for which $g(x) = 3$, or indicate there are no such values.

(ii) Determine the end behavior of $g$ as $x$ decreases without bound. Express your answer using the mathematical notation of a limit.

(C) (i) Use the table of values of $f(x)$ to determine if $f$ is best modeled by a linear, quadratic, exponential, or logarithmic function.

(ii) Give a reason for your answer based on the relationship between the change in the output values of $f$ and the change in the input values of $f$.

▶️ Answer/Explanation

Detailed solution

(A) (i)
First, evaluate the inner function $f(5)$ using the table: $f(5) = 34$.
Next, substitute this value into $g(x)$ to find $h(5) = g(34)$.
$g(34) = \frac{34^3 – 14(34) – 27}{34 + 2}$
$g(34) = \frac{39304 – 476 – 27}{36} = \frac{38801}{36}$
$h(5) \approx 1077.806$

(A) (ii)
To find $f^{-1}(4)$, we look for the input value $x$ in the table that produces an output of $4$.
The table shows that $f(3) = 4$.
Therefore, $f^{-1}(4) = 3$.

(B) (i)
Set $g(x) = 3$ and solve for $x$:
$\frac{x^3 – 14x – 27}{x + 2} = 3$
Multiply both sides by $(x+2)$: $x^3 – 14x – 27 = 3(x + 2)$
Simplify: $x^3 – 14x – 27 = 3x + 6$
Rearrange into polynomial form: $x^3 – 17x – 33 = 0$
Using a graphing calculator to find the zero of this polynomial:
$x \approx 4.879$

(B) (ii)
We evaluate the limit as $x \to -\infty$ for $g(x)$.
$\lim_{x \to -\infty} \frac{x^3 – 14x – 27}{x + 2}$
By examining the leading terms, the function behaves like $\frac{x^3}{x} = x^2$ for large absolute values of $x$.
As $x \to -\infty$, $x^2 \to \infty$.
$\lim_{x \to -\infty} g(x) = \infty$

(C) (i)
Calculate the first differences of $f(x)$: $(-5) – (-10) = 5$, $4 – (-5) = 9$, $17 – 4 = 13$, $34 – 17 = 17$.
Calculate the second differences: $9 – 5 = 4$, $13 – 9 = 4$, $17 – 13 = 4$.
Since the second differences are constant, $f$ is best modeled by a quadratic function.

(C) (ii)
The model is quadratic because for equal intervals of the input values $x$ (step size of $1$), the rate of change of the output values increases by a constant amount (constant second difference of $4$).

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions - Effective Fall 2023