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AP Precalculus -1.8 Rational Functions and Zeros- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.8 Rational Functions and Zeros- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.8 Rational Functions and Zeros- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question 

The figure shows the graph of the decreasing function \( f \) on its domain of all real numbers where \( x \neq 1 \). The points \( (-1, 0) \) and \( (3, 2) \) are on the graph of \( f \). The function \( g \) is given by \( g(x) = 2 + 3\ln x \).
(A) (i) The function \( h \) is defined by \( h(x) = (g \circ f)(x) = g(f(x)) \). Find the value of \( h(3) \) as a decimal approximation, or indicate that it is not defined.
       (ii) Find all real zeros of \( f \), or indicate there are no real zeros.
(B) (i) Find all values of \( x \), as decimal approximations, for which \( g(x) = e \), 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.
(C) (i) Determine if \( f \) is invertible.
       (ii) Give a reason for your answer based on the definition of a function and the graph of \( y = f(x) \).
▶️ Answer/Explanation
Detailed solution

(A)(i) Find \( h(3) \)
The function is defined as \( h(x) = g(f(x)) \).
First, we find the value of the inner function, \( f(3) \).
According to the problem text and graph, the point \( (3, 2) \) lies on the graph of \( f \). Therefore, \( f(3) = 2 \).
Now, substitute this value into the outer function \( g(x) \):
\( h(3) = g(2) \)
Using the definition \( g(x) = 2 + 3\ln x \):
\( g(2) = 2 + 3\ln(2) \)
Using a calculator to approximate \( \ln(2) \approx 0.693 \):
\( g(2) = 2 + 3(0.6931…) \approx 2 + 2.079 = 4.079 \)
Answer: \( h(3) \approx 4.079 \)

(A)(ii) Find real zeros of \( f \)
A real zero of a function occurs where the graph intersects the x-axis (where \( f(x) = 0 \)).
Looking at the provided graph, the curve intersects the x-axis at \( x = -1 \).
The problem text confirms the point \( (-1, 0) \) is on the graph.
There are no other intersections with the x-axis shown.
Answer: \( x = -1 \)

(B)(i) Find \( x \) for \( g(x) = e \)
Set up the equation:
\( 2 + 3\ln x = e \)
Subtract 2 from both sides:
\( 3\ln x = e – 2 \)
Divide by 3:
\( \ln x = \frac{e – 2}{3} \)
Convert from logarithmic to exponential form to solve for \( x \):
\( x = e^{\left(\frac{e – 2}{3}\right)} \)
Approximating the value (using \( e \approx 2.718 \)):
\( x \approx e^{0.2394} \)
Answer: \( x \approx 1.271 \)

(B)(ii) End behavior of \( g \)
We need to evaluate the limit as \( x \to \infty \) for \( g(x) = 2 + 3\ln x \).
As \( x \) increases without bound (\( x \to \infty \)), the natural logarithm function \( \ln x \) also increases without bound (\( \ln x \to \infty \)).
Multiplying by 3 and adding 2 does not change the unbounded nature.
Answer: \( \lim_{x \to \infty} g(x) = \infty \)

(C)(i) Is \( f \) invertible?
Answer: Yes, \( f \) is invertible.

(C)(ii) Reason
A function is invertible if and only if it is one-to-one. This can be verified visually using the Horizontal Line Test.
Looking at the graph of \( f(x) \):
1. The function is strictly decreasing on both branches of its domain (\( x < 1 \) and \( x > 1 \)).
2. The range of the left branch appears to be \( (-\infty, 1) \) and the range of the right branch appears to be \( (1, \infty) \).
Because the y-values do not repeat (no horizontal line intersects the graph more than once), the function is one-to-one.
Therefore, \( f \) has an inverse.

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