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

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