AP Precalculus -1.5 Polynomial Functions and Complex Zeros- FRQ Exam Style Questions - Effective Fall 2023
AP Precalculus -Link- FRQ Exam Style Questions – Effective Fall 2023
AP Precalculus -Link- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
Question
(i) The function \( h \) is defined by \( h(x) = (g \circ f)(x) = g(f(x)) \). Find the value of \( h(1) \) as a decimal approximation, or indicate that it is not defined. Show the work that leads to your answer.
(ii) Find the value of \( f^{-1}(3.5) \), or indicate that it is not defined.
(i) Find all values of \( x \), as decimal approximations, for which \( g(x) = 0 \), 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.
(i) Based on the table, which of the following function types best models function \( f \): linear, quadratic, exponential, or logarithmic?
(ii) Give a reason for your answer in part C (i) based on the relationship between the change in the output values of \( f \) and the change in the input values of \( f \). Refer to the values in the table in your reasoning.
Most-appropriate topic codes (AP Precalculus CED):
• 1.6: Polynomial Functions and End Behavior – part B(ii)
• 2.3: Exponential Functions – part C(i), C(ii)
• 2.7: Composite Functions – part A(i)
• 2.8: The Inverse of a Function – part A(ii)
▶️ Answer/Explanation
A (i)
We have \( h(1) = g(f(1)) \).
From the table, \( f(1) = 1.75 \).
Then \( g(1.75) = -0.167(1.75)^3 + (1.75)^2 – 1.834 \).
Compute step by step:
\( (1.75)^3 = 5.359375 \)
\( -0.167 \times 5.359375 \approx -0.895 \) (using three decimal places for intermediate steps)
\( (1.75)^2 = 3.0625 \)
So, \( g(1.75) \approx -0.895 + 3.0625 – 1.834 = 0.3335 \).
Rounded to three decimal places: \( 0.333 \).
✅ Answer: \(\boxed{0.333}\)
A (ii)
We want \( f^{-1}(3.5) \), which is the input value \( x \) such that \( f(x) = 3.5 \).
From the table, when \( x = 0 \), \( f(x) = 3.5 \).
Thus, \( f^{-1}(3.5) = 0 \).
✅ Answer: \(\boxed{0}\)
B (i)
Solve \( g(x) = 0 \): \( -0.167x^3 + x^2 – 1.834 = 0 \).
Using a graphing calculator (as permitted in the original problem context):
• Graph \( y = -0.167x^3 + x^2 – 1.834 \) and find the x‑intercepts.
• Alternatively, use the calculator’s equation solver.
The three real roots (to three decimal places) are:
\( x \approx -1.233 \), \( x \approx 1.578 \), \( x \approx 5.643 \).
✅ Answer: \(\boxed{-1.233, \; 1.578, \; 5.643}\)
B (ii)
For a polynomial, end behavior is determined by the leading term.
For \( g(x) = -0.167x^3 + x^2 – 1.834 \), the leading term is \( -0.167x^3 \).
As \( x \to \infty \), \( x^3 \to \infty \), so \( -0.167x^3 \to -\infty \).
Thus, \( \lim_{x \to \infty} g(x) = -\infty \).
✅ Answer: \(\boxed{\lim_{x \to \infty} g(x) = -\infty}\)
C (i)
Examine the table: As \( x \) increases by 1 each time, \( f(x) \) is halved.
This constant multiplicative rate of change (ratio) indicates an exponential decay model.
✅ Answer: \(\boxed{\text{Exponential}}\)
C (ii)
Reasoning: For equal-length input intervals of 1, the ratios of successive output values are constant:
\( \frac{f(-1)}{f(-2)} = \frac{7}{14} = 0.5 \),
\( \frac{f(0)}{f(-1)} = \frac{3.5}{7} = 0.5 \),
\( \frac{f(1)}{f(0)} = \frac{1.75}{3.5} = 0.5 \),
\( \frac{f(2)}{f(1)} = \frac{0.875}{1.75} = 0.5 \).
Since the output values change by a constant factor (0.5) over equal input intervals, the data are best modeled by an exponential function.
✅ Scoring Note: The explanation must reference the constant ratio over equal input intervals.
▶️ Answer/Explanation
(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, \( f(-3)=f(0)=f(3)=1 \).
▶️ Answer/Explanation
a. Graphing the function
Using a graphing utility for \(P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888\) on the interval \([0, 200]\) reveals a cubic curve shape.
The graph starts with a slight dip to a local minimum near \(x=0\), then rises steeply to a local maximum near \(x=180\), before falling again.
b. Average rate of change between extrema
First, find the derivative: \(P'(x) = -0.0039x^2 + 0.7014x – 0.4591\).
Set \(P'(x) = 0\) and use the quadratic formula to find the extrema: \(x \approx 0.66\) (local min) and \(x \approx 179.19\) (local max).
Calculate the profit at these points: \(P(0.66) \approx -422.04\) and \(P(179.19) \approx 3276.57\).
The average rate of change is: \(\frac{3276.57 – (-422.04)}{179.19 – 0.66} = \frac{3698.61}{178.53} \approx 20.72\).
c. Equation of the secant line
The slope \(m\) was found in part (b) to be approximately \(20.72\).
Using the point-slope form \(y – y_1 = m(x – x_1)\) with the minimum point \((0.66, -422.04)\):
\(y – (-422.04) = 20.72(x – 0.66)\)
\(y = 20.72x – 13.68 – 422.04\)
The equation is approximately \(y = 20.72x – 435.72\).
d. Inflection point
Find the second derivative: \(P”(x) = -0.0078x + 0.7014\).
Set \(P”(x) = 0\) to find the change in concavity: \(0 = -0.0078x + 0.7014 \Rightarrow x = \frac{0.7014}{0.0078} \approx 89.92\).
Find the corresponding y-value: \(P(89.92) \approx 1427.27\).
The inflection point is approximately \((89.92, 1427.27)\).
e. Variation of rate of change
The rate of change is represented by the derivative, \(P'(x)\).
Before the inflection point (\(x < 89.92\)), the graph is concave up (\(P”(x) > 0\)), so the rate of change is increasing.
After the inflection point (\(x > 89.92\)), the graph is concave down (\(P”(x) < 0\)), so the rate of change is decreasing.