AP Precalculus -1.7 Rational Functions and End Behavior- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.7 Rational Functions and End Behavior- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.7 Rational Functions and End Behavior- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

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 \(f\).

Most-appropriate topic codes (AP Precalculus CED):

• 2.7: Composition of Functions – part A(i)
• 2.8: Inverse Functions – part A(ii)
• 1.8: Rational Functions and Zeros – part B(i)
• 1.7: Rational Functions and End Behavior – part B(ii)
• 1.13: Function Model Selection and Assumption Articulation – part C

▶️ 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 \(4\)).

Question

The figure shows the graph of the function \( f \) on its domain of \( -3.5 \le x \le 3.5 \). The points \( (-3,1) \), \( (0,1) \), and \( (3,1) \) are on the graph of \( f \). The function \( g \) is given by \( g(x)=2.916\cdot(0.7)^x \).

(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 values of \( x \) for which \( f(x)=1 \), or indicate that there are no such values.

(B)
(i) Find all values of \( x \), as decimal approximations, for which \( g(x)=2 \), or indicate that 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 \( f \) has an inverse function.
(ii) Give a reason for your answer based on the definition of a function and the graph of \( y=f(x) \).

Most-appropriate topic codes (CED):

• TOPIC 1.7: Composition of Functions — part (A)
• TOPIC 2.3: Exponential Functions — part (B)
• TOPIC 1.5: Inverse Functions — part (C)

▶️ Answer/Explanation

Concise solution

(A)(i)
From the graph, \( f(3)=1 \).
\( h(3)=g(f(3))=g(1)=2.916(0.7)=2.041 \).

(A)(ii)
From the graph, \( f(x)=1 \) at \( x=-3,0,3 \).

(B)(i)
\( 2.916(0.7)^x=2 \)
\( (0.7)^x=\frac{2}{2.916} \)
Taking logarithms,
\( x=\frac{\ln(\frac{2}{2.916})}{\ln(0.7)}\approx1.057 \).

(B)(ii)
Since \( 0.7<1 \), \( (0.7)^x \to 0 \) as \( x \to \infty \).
\( \lim_{x\to\infty} g(x)=0 \).

(C)
\( f \) does not have an inverse because it is not one-to-one.
For example, \( f(-3)=f(0)=f(3)=1 \).

Question

The figure shows the graph of the increasing function \( f \) on its domain of all real numbers \( x > 2 \). The points \( (3, 0) \) and \( (6, 3) \) are on the graph of \( f \). The function \( g \) is given by \( g(x) = \frac{9}{(x-3)} \).

(A) (i) The function \( h \) is defined by \( h(x) = (g \circ f)(x) = g(f(x)) \). Find the value of \( h(6) \) 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 \( g(x) = -5.8 \), or indicate that 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) Determine if \( f \) is invertible.

(ii) Give a reason for your answer based on the definition of a function and the graph of \( y = f(x) \).

Most-appropriate topic codes (AP Precalculus CED):

• 2.7: Composition of Functions — part (A)(i)
• 1.8: Rational Functions and Zeros — parts (A)(ii) and (B)(i)
• 1.7: Rational Functions and End Behavior — part (B)(ii)
• 2.8: Inverse Functions — parts (C)(i) and (C)(ii)

▶️ Answer/Explanation

(A)

(i) Value of \( h(6) \)
The composition is \( h(6) = g(f(6)) \).
From the graph, the point \( (6, 3) \) is on \( f \), so \( f(6) = 3 \).
Then \( h(6) = g(3) = \frac{9}{(3-3)} = \frac{9}{0} \), which is undefined.
✅ Answer: \( \boxed{\text{Not defined}} \)

(ii) Real zeros of \( f \)
A zero of \( f \) is an \( x \)-value such that \( f(x) = 0 \).
From the given information, the point \( (3, 0) \) is on the graph of \( f \). Therefore, \( f(3) = 0 \).
✅ Answer: \( \boxed{3} \)

(B)

(i) Solving \( g(x) = -5.8 \)
Set \( g(x) = \frac{9}{x-3} = -5.8 \).
Solve for \( x \):
\( 9 = -5.8(x – 3) \)
\( 9 = -5.8x + 17.4 \)
\( -8.4 = -5.8x \)
\( x = \frac{-8.4}{-5.8} \approx 1.448275… \)
✅ Answer (decimal approximation): \( \boxed{1.448} \)

(ii) End behavior as \( x \) decreases without bound
For \( g(x) = \frac{9}{x-3} \), as \( x \to -\infty \), the denominator \( x-3 \to -\infty \).
A constant numerator divided by a value approaching \(-\infty\) approaches 0.
✅ Answer (limit notation): \( \boxed{\lim_{x \to -\infty} g(x) = 0} \)

(C)

(i) & (ii) Invertibility of \( f \)
A function is invertible if and only if it is one-to-one (each output comes from exactly one input).
The problem states that \( f \) is increasing on its entire domain \( x > 2 \). An increasing function is always one-to-one, because if \( a \neq b \), then either \( a < b \) (so \( f(a) < f(b) \)) or \( a > b \) (so \( f(a) > f(b) \)), meaning \( f(a) \neq f(b) \).
Since \( f \) is one-to-one on its domain, it is invertible.
✅ Answer: \( \boxed{\text{Yes}} \)

AP Precalculus -1.7 Rational Functions and End Behavior- FRQ Exam Style Questions - Effective Fall 2023