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AP Calculus BC 2.8 The Product Rule Exam Style Questions – FRQ

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

\(G'(c) = \frac{8}{3}\).

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