AP Calculus BC 2.8 The Product Rule Exam Style Questions - FRQ
Question
(a) Topic-5.6- Determining Concavity of Functions over Their Domains
(b) Topic-2.8- The Product Rule
(c) Topic-4.7- Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
(d) Topic-5.1- Using the Mean Value Theorem
Let \(f\) be a continuous function defined on the closed interval \(−4 \leq x \leq 6\). The graph of \(f\), consisting of four line segments, is shown above. Let \(G\) be the function defined by \(G(x)=\int_{0}^{x}f(t)dt.\)
(a) On what open intervals is the graph of \(G\) concave up? Give a reason for your answer.
(b) Let \(P\) be the function defined by \(P (x) = G (x)\cdot f(x)\). Find \(P'(3 )\).
(c) Find \(\lim_{x\rightarrow 2}\frac{G(x)}{x^{2}-2x}.\)
(d) Find the average rate of change of G on the interval \([−4, 2]\). Does the Mean Value Theorem guarantee a value \(c\), \(−4 < c < 2\), for which \(G'(c)\) is equal to this average rate of change? Justify your answer.
▶️Answer/Explanation
4(a)
\(G'(x) = f(x)\)
The graph of \(G\) is concave up for \( -4 < x < -2\) and \(2 < x < 6\), because \(G’ = f\) is increasing on these intervals.
4(b)
\(P'(x) = G(x) \cdot f'(x) + f(x) \cdot G'(x)\)
\(P'(3) = G(3) \cdot f'(3) + f(3) \cdot G'(3)\)
Substituting \(G(3) = \int_{0}^{3} f(t)\), \(dt = -3.5\) and \(G'(3) = f(3) = -3\) into the above expression for \(P'(3)\) gives the following:
\(P'(3) = -3.5 \cdot 1 + (-3) \cdot (-3) = 5.5\)
4(c)
\(\lim_{x \to 2} \left(x^2 – 2x\right) = 0\)
Because \(G\) is continuous for \(-4 \leq x \leq 6\), \(\lim_{x \to 2} G(x) = \int_{0}^{2} f(t)\), \(dt = 0\).
Therefore, the limit \(\lim_{x \to 2} \frac{G(x)}{x^2 – 2x}\) is an indeterminate form of type \(\frac{0}{0}\).
Using L’Hopital’s Rule,
\(\lim_{x \to 2} \frac{G(x)}{x^2 – 2x} = \lim_{x \to 2} \frac{G'(x)}{2x – 2}\)
\(= \lim_{x \to 2} \frac{f(x)}{2x – 2} = \frac{f(2)}{2} = \frac{-4}{2} = -2\)
4(d)
\(G(2) = \int_{0}^{2} f(t)\), \(dt = 0 \quad\) and \(\quad G(-4) = \int_{0}^{-4} f(t)\), \(dt = -16\)
Average rate of change \(= \frac{G(2) – G(-4)}{2 – (-4)} = \frac{0 – (-16)}{6} = \frac{8}{3}\)
Yes, \(G'(x) = f(x)\) so \(G\) is differentiable on \((-4, 2)\) and continuous on \([-4, 2]\).
Therefore, the Mean Value Theorem applies and guarantees a value \(c\), \(-4 < c < 2\), such that
\(G'(c) = \frac{8}{3}\).