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AP Precalculus -1.14 Function Models: Construction and Application- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.14 Function Models: Construction and Application- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.14 Function Models: Construction and Application- 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 

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 \(f\).
▶️ 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 \(4\)).

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