Home / AP Calculus BC: 6.1 Exploring Accumulations of  Change – Exam Style questions with Answer- FRQ

AP Calculus BC: 6.1 Exploring Accumulations of  Change – Exam Style questions with Answer- FRQ

Question

(a) Topic-6.1-Exploring Accumulations of Change

(b) Topic-5.2-Extreme Value Theorem Global Versus Local Extrema and Critical Points

(c) Topic-3.6-Calculating Higher-Order Derivatives

4. The graph of the differentiable function f , shown for \(-6\leq x\leq 7\), has a horizontal tangent at x = −2 and is linear for \(0\leq x\leq 7\). Let R be the region in the second quadrant bounded by the graph of f , the vertical line x = −6, and the x- and y-axes. Region R has area 12.

(a) The function g is defined by \(g(x)=\int_{0}^{x}f(x)dt\). Find the values of g(−6), g(4 ) , and g( 6) .

(b) For the function g defined in part (a), find all values of x in the interval \(0\leq x\leq 6\) at which the graph of g has a critical point. Give a reason for your answer.

(c) The function h is defined by \(h(x)=\int_{-6}^{0}f(t)dt\). Find the values of h( 6) , h'(6 ) , and h”( 6) . Show the work that leads to your answers.

▶️Answer/Explanation

4(a) The function g is defined by \(g(x)=\int_{0}^{x}f(t)dt\). Find the values of g(—6), g(4), and g(6).

\(g(-6)=\int_{0}^{-6}f(t)dt=-\int_{-6}^{0}f(t)dt=-12\)

\(g(6)=\int_{0}^{4}f(t)dt=\frac{1}{2}.4.2=4\)

\(g(6) =\int_{0}^{6}f(t)dt=\frac{1}{2}.4.2-\frac{1}{2}.2.1=3\)

4(b) For the function g defined in part (a), find all values of x in the interval\(0\leq x\leq 6\)   at which the graph of g has a critical point. Give a reason for your answer.

g'(x) = f(x)

\(g'(x)=f(x)=0\Rightarrow x=4\)

Therefore, the graph of g has a critical point at x = 4.

4(c) The function A is defined by\(h(x)=\int_{-6}^{x}f(t)dt\) . Find the values of h(6), 4′(6), and h”(6). Show the work that leads to your answers.

\(h(6)=\int_{-6}^{6}f'(t)dt=f(6)-f(-6)=-1-0.5=-1.5\)

\(h'(x)=f'(x),so\\\ h'(6)=f'(6)=-\frac{1}{2}\)

\(h”(x)=f”(x),so\\\ h”(6)=f”(6)=0\)

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