AP Precalculus -2.11 Logarithmic Functions- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.11 Logarithmic Functions- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.11 Logarithmic Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question

The domain of $f$ consists of the five real numbers 1, 2, 3, 4, and 5. The table defines the function $f$ for these values.

The function $g$ is given by $g(x)=2 \ln x$.

(A)
(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.

(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 $x$ 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 table of values of $f(x)$.

Most-appropriate topic codes (AP Precalculus CED):

• 2.7: Composition of Functions – part A(i)
• 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.

Question

Students took a math review course to prepare them for college. At the beginning of the course, they were given a pretest to see how much they remembered. At that time $(t = 0)$ the average score was 60 out of 100. The students were then evaluated again at the end of each week. After 4 weeks $(t = 4)$ the average score had increased to 81 out of 100.

The average score can be modeled by $S(t) = a + b\ln (t + 1)$ , where $S(t)$ is the average score for month $t$ , and $t$ is the number of weeks since the course began.

(A)
(i) Use the given data to write two equations that can be used to find the values for constants $a$ and $b$ in the expression $S(t)$.
(ii) Find the values for $a$ and $b$.

(B)
(i) Use the given data to find the average rate of change of the scores in points per week, from $t = 0$ to $t = 4$ weeks. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Interpret the meaning of your answer from (i) in the context of the problem.
(iii) Consider the average rates of change of $S$ from $t = 4$ to $t = p$ weeks, where $p > 4$. Are these average rates of change less than or greater than the average rate of change from $t = 0$ to $t = 4$ months found in (i)? Explain your reasoning.

(C) Explain how the domain values should be limited by the context of the problem.

Most-appropriate topic codes (AP Precalculus CED):

• 2.14: Logarithmic Function Context and Data Modeling – part A
• 1.2: Rates of Change – part B
• 2.11: Logarithmic Functions – part B(iii)
• 1.13: Function Model Selection and Assumption Articulation – part C

▶️ Answer/Explanation

A (i)
Given $S(0) = 60$ and $S(4) = 81$. The equations are: \[ a + b\ln(0+1) = 60 \] \[ a + b\ln(4+1) = 81 \]
✅ Answer: $a + b\ln(1) = 60$ and $a + b\ln(5) = 81$.

A (ii)
Since $\ln(1) = 0$: From the first equation, $a = 60$. Substituting into the second: $60 + b\ln(5) = 81 \implies b\ln(5) = 21$. \[ b = \frac{21}{\ln 5} \approx \frac{21}{1.6094} \approx 13.048 \]
✅ Answer: $a = 60$, $b \approx 13.048$.

B (i)
The average rate of change from $t=0$ to $t=4$: \[ \frac{S(4) – S(0)}{4 – 0} = \frac{81 – 60}{4} = \frac{21}{4} = 5.25 \]
✅ Answer: $5.25$ points per week.

B (ii)
The value 5.25 represents the average increase in the group’s score per week over the first 4 weeks.
✅ Answer: On average, the students’ scores increased by 5.25 points per week over the first 4 weeks.

B (iii)
The function $S(t) = a + b\ln(t+1)$ with $b > 0$ is a logarithmic function. Logarithmic functions are concave down. The average rate of change of a concave down function decreases as the input interval shifts to larger values.
✅ Answer: The average rates of change for $t = 4$ to $t = p$ (where $p > 4$) will be less than the AROC from $t=0$ to $t=4$, because the logarithmic model is concave down, resulting in decreasing rates of change.

C
Time in weeks cannot be negative. Additionally, the context implies a maximum score of 100 on the test. The model predicts scores exceeding 100 for large $t$ (e.g., $t > 20.4$), which is impossible for the context.
✅ Answer: The domain should be limited to non-negative weeks up to the point where the score reaches 100. (Example: $0 \le t \le 20$).

Question

An increasing, continuous function $g(x)$ has outputs of some values of $x$ shown in the table. $f(x)$ 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 $j(x) = \frac{1}{4.998} \ln(x) \cdot e^{2\sin(\sqrt{x})}$ could be used to model $f(x)$.
(i) Determine the end behavior of $j$ when $x$ 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 $x$ 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 (AP Precalculus CED):

• 1.9: Rational Functions and Vertical Asymptotes – part A(ii)
• 2.3: Exponential Functions – part B(i), B(ii)
• 2.8: Inverse Functions – part A(i)
• 2.11: Logarithmic Functions – part B(i), B(ii), C
• 2.14: Logarithmic Function Context and Data Modeling – part C

▶️ Answer/Explanation

Detailed solution

(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.

Question

(A) The functions $ g $ and $ h $ are given by
\[ g(x) = 2 \log_3 x \quad]
[\quad h(x) = 4 \cos^2 x. \]
(i) Solve $ g(x) = 4 $ for values of $ x $ in the domain of $ g $.
(ii) Solve $ h(x) = 3 $ for values of $ x $ in the interval $[0, \frac{\pi}{2})$.

(B) The functions $ j $ and $ k $ are given by
\[ j(x) = \log_2 x + 3 \log_2 2 \quad \text{and} \quad k(x) = \frac{6}{\tan x (\csc^2 x – 1)}. \]
(i) Rewrite $ j(x) $ as a single logarithm base 2 without negative exponents in any part of the expression. Your result should be of the form $ \log_2 (\text{expression}) $.
(ii) Rewrite $ k(x) $ as an expression in which $ \tan x $ appears exactly once and no other trigonometric functions are involved.

(C) The function $ m $ is given by $ m(x) = e^{2x} – e^x – 12 $. Find all input values in the domain of $ m $ that yield an output value of 0.

Most-appropriate topic codes (AP Precalculus CED):

• 2.11: Logarithmic Functions — part A(i), B(i)
• 2.13: Exponential and Logarithmic Equations and Inequalities — part A(i), C
• 3.9: Trigonometric Equations and Inequalities — part A(ii)
• 3.4: Trigonometric Identities — part B(ii)

▶️ Answer/Explanation

A. (i) Solving $ g(x) = 4 $:
Given $ g(x) = 2\log_3 x $. Set $ g(x) = 4 $: \[ 2\log_3 x = 4 \] Divide both sides by 2: \[ \log_3 x = 2 \] Convert to exponential form: \[ x = 3^2 = 9 \] Check domain: $ x > 0 $, so $ x = 9 $ is valid. ✅ Answer: $\boxed{x = 9}$

A. (ii) Solving $ h(x) = 3 $ on $ [0, \frac{\pi}{2}) $:
Given $ h(x) = 4\cos^2 x $. Set $ h(x) = 3 $: \[ 4\cos^2 x = 3 \] Divide both sides by 4: \[ \cos^2 x = \frac{3}{4} \] Take square root: \[ \cos x = \pm \frac{\sqrt{3}}{2} \] On $ [0, \frac{\pi}{2}) $, cosine is positive, so \[ \cos x = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6} \] ✅ Answer: $\boxed{x = \frac{\pi}{6}}$

B. (i) Rewriting $ j(x) $ as a single logarithm:
Given $ j(x) = \log_2 x + 3\log_2 2 $.
Use power rule: $ 3\log_2 2 = \log_2 (2^3) = \log_2 8 $.
Use product rule: \[ j(x) = \log_2 x + \log_2 8 = \log_2 (8x) \] ✅ Answer: $\boxed{\log_2(8x)}$

B. (ii) Rewriting $ k(x) $ in terms of $ \tan x $ only:
Given $ k(x) = \dfrac{6}{\tan x (\csc^2 x – 1)} $.
Use Pythagorean identity: $ \csc^2 x – 1 = \cot^2 x $.
Thus, \[ k(x) = \frac{6}{\tan x \cdot \cot^2 x} \] Since $ \cot x = \frac{1}{\tan x} $, we have $ \cot^2 x = \frac{1}{\tan^2 x} $.
Substitute: \[ k(x) = \frac{6}{\tan x \cdot \frac{1}{\tan^2 x}} = \frac{6}{\frac{\tan x}{\tan^2 x}} = \frac{6}{\frac{1}{\tan x}} = 6 \tan x \] ✅ Answer: $\boxed{6 \tan x}$

C. Solving $ m(x) = 0 $:
Given $ m(x) = e^{2x} – e^x – 12 $.
Set $ m(x) = 0 $: \[ e^{2x} – e^x – 12 = 0 \] Let $ y = e^x $. Then $ e^{2x} = (e^x)^2 = y^2 $.
Substitute: \[ y^2 – y – 12 = 0 \] Factor: \[ (y – 4)(y + 3) = 0 \] Thus $ y = 4 $ or $ y = -3 $.
Since $ y = e^x > 0 $, discard $ y = -3 $.
So $ e^x = 4 $.
Take natural log: $ x = \ln 4 $.
✅ Answer: $\boxed{x = \ln 4}$

Question

On the initial day of sales ($ t=0 $) for a new video game, there were $ 40 $ thousand units of the game sold that day. Ninety-one days later ($ t=91 $), there were $ 76 $ thousand units of the game sold that day.

The number of units of the video game sold on a given day can be modeled by the function $ G(t)=\ln(a+bt+1) $, where $ G(t) $ is the number of units sold, in thousands, on day $ t $ since the initial day of sales.

(A)
(i) Use the given data to write two equations that can be used to find the values for constants $ a $ and $ b $ in the expression for $ G(t) $.
(ii) Find the values for $ a $ and $ b $ as decimal approximations.

(B)
(i) Use the given data to find the average rate of change of the number of units of the video game sold, in thousands per day, from $ t=0 $ to $ t=91 $ days. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Use the average rate of change found in part (i) to estimate the number of units of the video game sold, in thousands, on day $ t=50 $. Show the work that leads to your answer.
(iii) Let $ A_t $ represent the estimate of the number of units of the video game sold, in thousands, using the average rate of change found in part (i). For $ A_{50} $, found in part (ii), it can be shown that $ A_{50} < G(50) $. Explain why, in general, $ A_t < G(t) $ for all $ 0<t<91 $.

(C)
The makers of the video game reported that daily sales of the video game decreased each day after $ t=91 $. Explain why the error in the model $ G $ increases after $ t=91 $.

Most-appropriate topic codes (CED):

• TOPIC 2.11: Logarithmic Models — part (A)
• TOPIC 1.2: Average Rate of Change — part (B)
• TOPIC 1.4: Function Behavior — part (C)

▶️ Answer/Explanation

Concise solution

(A)(i)
$ G(0)=\ln(a+1)=40 $
$ G(91)=\ln(a+91b+1)=76 $

(A)(ii)
From $ \ln(a+1)=40 \Rightarrow a+1=e^{40} $.
From $ \ln(a+91b+1)=76 \Rightarrow a+91b+1=e^{76} $.
Subtracting gives $ 91b=e^{76}-e^{40} \Rightarrow b=\frac{e^{76}-e^{40}}{91} $.

(B)(i)
Average rate of change $=\frac{76-40}{91}=\frac{36}{91}\approx0.396 $.

(B)(ii)
$ A_{50}=40+0.396(50)=59.8 $.

(B)(iii)
$ G(t) $ is increasing and concave down, so the secant line from $ t=0 $ to $ t=91 $ lies below the graph for $ 0<t<91 $.

(C)
After $ t=91 $, actual sales decrease while the model $ G $ continues to increase, so the difference between actual values and model values grows.

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

Directions:

Unless otherwise specified, the domain of a function $ f $ is assumed to be the set of all real numbers $ x $ for which $ f(x) $ is a real number. Angle measures for trigonometric functions are assumed to be in radians.
Solutions to equations must be real numbers. Determine the exact value of any expression that can be obtained without a calculator. For example, $ \log_2 8 $, $ \cos(\frac{\pi}{2}) $, and $ \sin^{-1}(1) $ can be evaluated without a calculator.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms, where applicable. For example, $ 2x + 3x $, $ 5^2 \cdot 5^3 $, $ \frac{x^5}{x^2} $, and $ \ln 3 + \ln 5 $ should be rewritten in equivalent forms.
For each part of the question, show the work that leads to your answers.

Part A
The functions $ g $ and $ h $ are given by

$ g(x) = \log_5(4x – 2) $
$ h(x) = \sin^{-1}(8x) $

(i) Solve $ g(x) = 3 $ for values of $ x $ in the domain of $ g $.

(ii) Solve $ h(x) = \frac{\pi}{4} $ for values of $ x $ in the domain of $ h $.

Part B
The functions $ j $ and $ k $ are given by

$ j(x) = (\sec x)(\cot x) $
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $

(i) Rewrite $ j(x) $ as an expression involving $ \sin x $ and no other trigonometric functions.

(ii) Rewrite $ k(x) $ as an expression of the form $ 4^{(ax+b)} $, where $ a $ and $ b $ are constants.

Part C
The function $ m $ is given by

$ m(x) = \sqrt{3}\tan(x + \frac{\pi}{2}) $.

Find all values in the domain of $ m $ that yield an output value of 1.

▶️ Answer/Explanation

Detailed solution

Part A
(i) Solve $ g(x) = 3 $
Start with the given equation:
$ \log_5(4x – 2) = 3 $
Convert the logarithmic equation to exponential form ($ y = \log_b x \iff x = b^y $):
$ 4x – 2 = 5^3 $
Evaluate the exponent:
$ 4x – 2 = 125 $
Add 2 to both sides:
$ 4x = 127 $
Divide by 4:
$ x = \frac{127}{4} $

(ii) Solve $ h(x) = \frac{\pi}{4} $
Start with the given equation:
$ \sin^{-1}(8x) = \frac{\pi}{4} $
Take the sine of both sides to isolate the argument:
$ 8x = \sin\left(\frac{\pi}{4}\right) $
Substitute the exact value of $ \sin\left(\frac{\pi}{4}\right) $:
$ 8x = \frac{\sqrt{2}}{2} $
Divide by 8:
$ x = \frac{\sqrt{2}}{16} $

Part B
(i) Rewrite $ j(x) $
Start with the function definition:
$ j(x) = (\sec x)(\cot x) $
Substitute the reciprocal and quotient identities ($ \sec x = \frac{1}{\cos x} $ and $ \cot x = \frac{\cos x}{\sin x} $):
$ j(x) = \left(\frac{1}{\cos x}\right) \left(\frac{\cos x}{\sin x}\right) $
Cancel the $ \cos x $ terms:
$ j(x) = \frac{1}{\sin x} $

(ii) Rewrite $ k(x) $
Start with the function definition:
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $
To write in the form $ 4^{(ax+b)} $, convert bases 16 and 2 to base 4.
Since $ 16 = 4^2 $ and $ 2 = \sqrt{4} = 4^{1/2} = 4^{0.5} $:
$ k(x) = \frac{(4^2)^{3x} \cdot 4^x}{4^{0.5}} $
Apply the power of a power rule ($ (a^m)^n = a^{mn} $):
$ k(x) = \frac{4^{6x} \cdot 4^x}{4^{0.5}} $
Apply the product rule for exponents ($ a^m \cdot a^n = a^{m+n} $) in the numerator:
$ k(x) = \frac{4^{6x + x}}{4^{0.5}} = \frac{4^{7x}}{4^{0.5}} $
Apply the quotient rule for exponents ($ \frac{a^m}{a^n} = a^{m-n} $):
$ k(x) = 4^{7x – 0.5} $ (or $ 4^{7x – \frac{1}{2}} $)
Thus, $ a = 7 $ and $ b = -0.5 $.

Part C
Find values where $ m(x) = 1 $
Set the function equal to 1:
$ \sqrt{3}\tan\left(x + \frac{\pi}{2}\right) = 1 $
Isolate the tangent function by dividing by $ \sqrt{3} $:
$ \tan\left(x + \frac{\pi}{2}\right) = \frac{1}{\sqrt{3}} $
Determine the reference angle. We know that $ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} $.
Set up the general solution for tangent ($ \theta = \text{ref} + n\pi $):
$ x + \frac{\pi}{2} = \frac{\pi}{6} + n\pi $, where $ n $ is any integer.
Solve for $ x $ by subtracting $ \frac{\pi}{2} $ from both sides:
$ x = \frac{\pi}{6} – \frac{\pi}{2} + n\pi $
Find a common denominator (6) to combine fractions:
$ x = \frac{\pi}{6} – \frac{3\pi}{6} + n\pi $
$ x = -\frac{2\pi}{6} + n\pi $
Simplify the fraction:
$ x = -\frac{\pi}{3} + n\pi $

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.

(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.

AP Precalculus -2.11 Logarithmic Functions- FRQ Exam Style Questions - Effective Fall 2023