AP Precalculus -2.8 Inverse Functions- FRQ Exam Style Questions - Effective Fall 2023
AP Precalculus -2.8 Inverse Functions- FRQ Exam Style Questions – Effective Fall 2023
AP Precalculus -2.8 Inverse Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
(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
(i) The function $h$ is defined by $h(x)=(g \circ f)(x)=g(f(x))$. Find the value of $h(4)$ as a decimal approximation, or indicate that it is not defined.
(ii) Find all values of $x$ for which $f(x)=3$, or indicate there are no such values.
(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 $x$ increases without bound. Express your answer using the mathematical notation of a limit.
(i) Determine if $f$ has an inverse function.
(ii) Give a reason for your answer based on the definition of a function and the table of values of $f(x)$.
Most-appropriate topic codes (AP Precalculus CED):
• 2.8: Inverse Functions – part A(ii), C
• 2.11: Logarithmic Functions – part B(ii)
• 2.13: Exponential and Logarithmic Equations and Inequalities – part B(i)
▶️ Answer/Explanation
A (i)
We need to find $h(4)=g(f(4))$.
From the table, the output for input $4$ is $f(4)=3$.
Substitute this into $g(x)$: $g(3)=2 \ln 3$.
Using a calculator, $2 \ln 3 \approx 2.197$.
✅ Answer: $\boxed{2.197}$
A (ii)
We are looking for the inputs where the output is 3.
From the table, $f(x)=3$ when $x=2$ and $x=4$.
✅ Answer: $\boxed{x=2, x=4}$
B (i)
Set the equation: $g(x)=3 \implies 2 \ln x=3$.
Isolate the natural log: $\ln x=1.5$.
Exponentiate both sides to solve for $x$: $x=e^{1.5} \approx 4.482$.
✅ Answer: $\boxed{4.482}$
B (ii)
The function $g$ is an increasing logarithmic function. As $x$ increases without bound, the natural log function grows slowly but infinitely, so $g(x)$ increases without bound.
Therefore, $\lim_{x \to \infty} g(x) = \infty$.
✅ Answer: $\boxed{\lim_{x \to \infty} g(x) = \infty}$
C (i)
$f$ does not have an inverse function on its domain of the five real numbers 1, 2, 3, 4, and 5.
✅ Answer: $\boxed{\text{No inverse function}}$
C (ii)
Reasoning: There are output values of $f$ that are not mapped from unique input values (the function $f$ is not one-to-one). Because $f(1)=1$ and $f(5)=1$, the inverse function on this domain does not exist. Note: Stating only that it “fails the horizontal line test” without referencing specific data values is generally insufficient for full credit.
▶️ Answer/Explanation
(A) i. \(g(f^{-1}(1))\)
- From the graph of \(f(x)\), \(f^{-1}(1)\) means find \(x\) such that \(f(x) = 1\).
- From the graph, \(f(x) = 1\) at approximately \(x \approx 2.5\). So \(f^{-1}(1) \approx 2.5\).
- From the table, \(g(2) = 1\), \(g(4) = 2\). Since \(g\) is increasing and continuous, by linear interpolation for \(x \approx 2.5\): \(g(2.5) \approx 1 + \frac{2.5-2}{4-2} \cdot (2-1) = 1 + 0.25 = 1.25\).
- Thus, \(g(f^{-1}(1)) \approx 1.25\).
(A) ii. \(h(x) = \frac{g(x)}{f(x)}\), \(x > 0\)
- Discontinuities occur where \(f(x) = 0\) or where \(f\) is undefined (but \(f\) appears continuous from the graph for \(x>0\)).
- From the graph, \(f(x) = 0\) at \(x \approx 1.5\) (where graph crosses x-axis). That is the only zero of \(f\) for \(x>0\).
- At \(x \approx 1.5\), \(g(x) \neq 0\) (since \(g\) is about 0.5 between \(g(1)=0\) and \(g(2)=1\)).
- Thus, \(h(x)\) has a vertical asymptote at \(x \approx 1.5\) because denominator → 0, numerator nonzero → \(h(x) \to \pm\infty\).
- Type: infinite discontinuity (vertical asymptote).
(B) i. End behavior of \(j(x) = \frac{1}{4.998} \ln(x) \cdot e^{2\sin(\sqrt{x})}\) as \(x \to \infty\)
- As \(x \to \infty\), \(\ln(x) \to \infty\).
- The factor \(e^{2\sin(\sqrt{x})}\) oscillates between \(e^{-2}\) and \(e^{2}\) because \(\sin(\sqrt{x})\) oscillates in \([-1,1]\).
- Thus \(j(x)\) oscillates with amplitude growing like \(\ln(x)\). The limit does not exist because oscillations persist and amplitude grows without bound.
- Limit expression: \(\lim_{x \to \infty} j(x)\) does not exist.
(B) ii. Vertical asymptote as \(x \to 0^+\)
- As \(x \to 0^+\), \(\ln(x) \to -\infty\).
- The factor \(e^{2\sin(\sqrt{x})} \to e^{0} = 1\) (since \(\sqrt{x} \to 0\), \(\sin(\sqrt{x}) \to 0\)).
- Thus \(j(x) \to -\infty\).
- Limit expression: \(\lim_{x \to 0^+} j(x) = -\infty\). So vertical asymptote at \(x=0\).
(C) Best model for \(g(x)\) from the table:
- Check changes in \(g(x)\) as \(x\) doubles:
- From \(x=0.5\) to \(x=1\), \(\Delta g = 1\).
- From \(x=1\) to \(x=2\), \(\Delta g = 1\).
- From \(x=2\) to \(x=4\), \(\Delta g = 1\).
- From \(x=4\) to \(x=8\), \(\Delta g = 1\).
- Each time \(x\) doubles, \(g(x)\) increases by about 1. This suggests a logarithmic relationship: \(g(x) \approx a + b \ln(x)\).
- Thus, \(g(x)\) is best modeled by a logarithmic function.
▶️ Answer/Explanation
(A)(i)
\( e^{2x}=10 \)
\( 2x=\ln(10) \)
\( x=\frac{1}{2}\ln(10) \).
(A)(ii)
\( \arcsin(x+3)=\frac{\pi}{4} \)
\( x+3=\sin\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \)
\( x=\frac{\sqrt{2}}{2}-3 \).
(B)(i)
\( j(x)=\log_{10}\!\left(\frac{8x^9}{2x^3}\right) \)
\( =\log_{10}(4x^6) \).
(B)(ii)
\( k(x)=\frac{\sin x}{\sec x}=\sin x\cos x \)
\( =\frac{\sin x}{\cos x}\cos^2 x=\tan x \).
(C)
\( \cos(\tan(2x-1))=0 \)
\( \tan(2x-1)=\frac{\pi}{2}+k\pi \)
\( 2x-1=\arctan\!\left(\frac{\pi}{2}+k\pi\right) \), where defined.
Question
Most-appropriate topic codes (AP Precalculus CED):
• 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: \( \boxed{\text{Yes}} \)
▶️ Answer/Explanation
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.
▶️ Answer/Explanation
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.
▶️ Answer/Explanation
(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.
▶️ Answer/Explanation
(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 \(4\)).