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AP Precalculus -2.3 Exponential Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.3 Exponential Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.3 Exponential Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

The value \( 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4.7 \) is the output value of an exponential function of the form \( f(x) = a \cdot b^x \), where \( a \) and \( b \) are constants. Which of the following describes the function and input value that corresponds to this output value?
(A) The exponential function has an initial value of 1 and a base of \( 2 \cdot 4.7 \), and the input value is 5.
(B) The exponential function has an initial value of 2 and a base of 2, and the input value is 4.7.
(C) The exponential function has an initial value of 4.7 and a base of 2, and the input value is 4.7.
(D) The exponential function has an initial value of 4.7 and a base of 2, and the input value is 5.
▶️ Answer/Explanation
Detailed solution

Given output value:
\(
2 \cdot 2 \cdot 2 \cdot 2 \cdot 4.7
\)

Step 1: Rewrite in exponential form
\(
2 \cdot 2 \cdot 2 \cdot 2 \cdot 4.7 = 4.7 \cdot 2^4
\)

Step 2: Compare with the exponential form
\(
f(x) = a \cdot b^x
\)

Comparing,
\(
a = 4.7,\quad b = 2,\quad x = 4
\)

Step 3: Match with the given options

The exponential function has initial value \(4.7\) and base \(2\).

\(
\boxed{\text{Correct answer: (D)}}
\)

Question 

The function \( f \) is given by \( f(x) = 5 \cdot (0.7)^x \). Which of the following describes \( f \)?
(A) The function \( f \) models exponential decay and \( \lim_{x \to \infty} f(x) = 0 \).
(B) The function \( f \) models exponential decay and \( \lim_{x \to -\infty} f(x) = \infty \).
(C) The function \( f \) models exponential growth and \( \lim_{x \to -\infty} f(x) = 0 \).
(D) The function \( f \) models exponential growth and \( \lim_{x \to \infty} f(x) = \infty \).
▶️ Answer/Explanation
Detailed solution

Base \( b = 0.7 \), which is between 0 and 1 ⇒ exponential decay.
As \( x \to \infty \), \( (0.7)^x \to 0 \) ⇒ \( f(x) \to 0 \).
As \( x \to -\infty \), \( (0.7)^x \to \infty \) ⇒ \( f(x) \to \infty \).
Option (A) matches decay and limit 0 at \( +\infty \).
✅ Answer: (A)

Question 

The exponential function \( f \) is defined by \( f(x) = ab^x \), where \( a \) and \( b \) are positive constants. The table gives values of \( f(x) \) at selected values of \( x \). Which of the following statements is true?
\[ \begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & \frac{3}{4} & \frac{3}{2} & 3 & 6 & 12 \\ \end{array} \]
(A) \( f \) demonstrates exponential decay because \( a > 0 \) and \( 0 < b < 1 \).
(B) \( f \) demonstrates exponential decay because \( a > 0 \) and \( b > 1 \).
(C) \( f \) demonstrates exponential growth because \( a > 0 \) and \( 0 < b < 1 \).
(D) \( f \) demonstrates exponential growth because \( a > 0 \) and \( b > 1 \).
▶️ Answer/Explanation
Detailed solution

From the table: \( f(0) = a = \frac{3}{4} > 0 \).
\( f(1) = ab = \frac{3}{2} \) ⇒ \( b = \frac{3/2}{3/4} = 2 \).
Since \( b = 2 > 1 \), \( f \) shows exponential growth.
✅ Answer: (D)

Question 

The table gives values of the function \( f \) for selected values of \( x \). Which of the following expressions could define \( f(x) \)?
\[ \begin{array}{c|cccc} x & 0 & 2 & 4 & 6 \\ \hline f(x) & 3 & 48 & 768 & 12{,}288 \\ \end{array} \]
(A) \( 2 + 4^x \)
(B) \( 3 \cdot 4^x \)
(C) \( 3 \cdot 16^x \)
(D) \( 4 \cdot 3^x \)
▶️ Answer/Explanation
Detailed solution

Check \( x=0 \): \( f(0) = 3 \), so \( a = 3 \) if \( f(x) = a \cdot b^x \).
Now check ratios: \( \frac{f(2)}{f(0)} = \frac{48}{3} = 16 \) ⇒ \( b^2 = 16 \) ⇒ \( b = 4 \) or \( -4 \) (positive base) ⇒ \( b=4 \).
Verify: \( 3 \cdot 4^2 = 3 \cdot 16 = 48 \) ✔
\( 3 \cdot 4^4 = 3 \cdot 256 = 768 \) ✔
\( 3 \cdot 4^6 = 3 \cdot 4096 = 12{,}288 \) ✔
Thus \( f(x) = 3 \cdot 4^x \).
✅ Answer: (B)

Question 

The function \( g \) has the property that for each time the input values double, the output values increase by 1. Which of the following could be the graph of \( y = g(x) \) in the \( xy \)-plane?

▶️ Answer/Explanation
Detailed solution

The property “when input doubles, output increases by 1” is characteristic of a logarithmic function with base 2: \( g(x) = \log_2 x \).
This function satisfies \( g(2x) = g(x) + 1 \).
Graphically, it is increasing, passes through (1,0), (2,1), (4,2), etc., with a vertical asymptote at \( x=0 \).
✅ Answer: (B) 

Question 

An exponential function \( G \) has a known common ratio of \(\frac{1}{2}\) and includes the input-output pair \((1,4)\). Which of the following could define \( G(x) \)?
(A) \( 4 + \frac{1}{2}(x – 1) \)
(B) \( \frac{1}{2} \cdot 4^x \)
(C) \( 4 \cdot \left( \frac{1}{2} \right)^x \)
(D) \( 4 \cdot \left( \frac{1}{2} \right)^{(x-1)} \)
▶️ Answer/Explanation
Detailed solution

Common ratio \( \frac{1}{2} \) means each term is multiplied by \( \frac{1}{2} \). Given \( G(1) = 4 \), we can write the exponential function in the form \( G(x) = a \cdot \left(\frac{1}{2}\right)^{x-k} \) where \( k \) is a shift.
Test (D): \( G(x) = 4 \cdot \left(\frac{1}{2}\right)^{(x-1)} \). Check \( x = 1 \): \( G(1) = 4 \cdot \left(\frac{1}{2}\right)^{0} = 4 \) ✅.
Check ratio: \( G(x+1) / G(x) = \frac{4 \cdot (1/2)^{x}}{4 \cdot (1/2)^{x-1}} = \frac{1}{2} \) ✅.
✅ Answer: (D)

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