Question
Topic-(a)- 6.4: The Fundamental Theorem of Calculus and Accumulation Functions
Topic-(b)-6.5: Interpreting the Behavior of Accumulation Functions Involving Area
Topic-(c)-6.4: The Fundamental Theorem of Calculus and Accumulation Functions
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 \(g\left ( x \right )=\int_{0}^{x}f\left ( t \right )dt\). 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 \(h\left ( x \right )=\int_{-6}^{x}f”\left ( t \right )dt\) . 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 \(g\left ( x \right )=\int_{0}^{x}f\left ( t \right )dt\) . Find the values of g(-6), g(4), and g(6).
\(g\left ( -6 \right )=\int_{0}^{-6}f\left ( t \right )dt=-\int_{-6}^{0}f\left ( t \right )dt=-12\)
\(g\left ( 4 \right )=\int_{0}^{4}f\left ( t \right )dt=\frac{1}{2}.4.2=4\)
\(g\left ( 6 \right )=\int_{0}^{6}f\left ( t \right )dt=\frac{1}{2}.4.2-\frac{1}{2}.2.1=3\)
4(b) For the function g defined in part (a), find all values of x in the interval 0 ≤ x ≤ 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 ⇒ x = 4
Therefore, the graph of g has a critical point at x = 4.
4(c) The function h is defined by h( x) = \(\int_{-6}^{x}f’\left ( t \right )dt\). Find the values of h(6), h'(6), and h”(6). Show the work that leads to your answers.
\(h\left ( 6 \right )=\int_{-6}^{6}f’\left ( t \right )dt=f\left ( 6 \right )-f\left ( -6 \right )=-1-0.5=-1.5\)
\(h’\left ( x \right )=f’\left ( x \right ), so h’\left ( 6 \right )=f\left ( 6 \right )=-\frac{1}{2}\).
\(h”\left ( x \right )=f”’\left ( x \right ), so h”\left ( 6 \right )=f”’\left ( 6 \right )=0\).