Question
(a) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(b) Topic-5.3 Determining Intervals on Which a Function is Increasing or Decreasing
(c) Topic-2.9 The Quotient Rule
(d) Topic-2.8 The Product Rule
The function f is defined on the closed interval [-5, 4] . The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by \(g(x)=\int_{-3}^{x}f(t)dt.\)
(a) Find g(3) .
(b) On what open intervals contained in -5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer.
(c) The function h is defined by \(h(x)= \frac{g(x)}{5x}\). Find h'(3).
(d) The function p is defined by p(x) = f(x2 – x). Find the slope of the line tangent to the graph of p at the point where x = -1.
▶️Answer/Explanation
\(
\textbf{3(a) } g(3) = \int_{-3}^{3} f(t) \, dt = 6 + 4 – 1 = 9
\)
\(
\textbf{3(b) } g'(x) = f(x)
\)
\(
\text{The graph of } g \text{ is increasing and concave down on the intervals } -5 < x < -3\)
\(\text{ and } 0 < x < 2 \text{ because } g’ = f \text{ is positive and decreasing on these intervals.}\)
\(
\textbf{3(c) } h'(x) = \frac{5x g'(x) – g(x) 5}{(5x)^2} = \frac{5x g'(x) – 5 g(x)}{25x^2}
\)
\(
h'(3) = \frac{(5)(3)g'(3) – 5g(3)}{25 \cdot 3^2}
\)
\(
= \frac{15(-2) – 5(9)}{225} = \frac{-75}{225} = -\frac{1}{3}
\)
\(
\textbf{3(d) } p'(x) = f'(x^2 – x)(2x – 1)
\)
\(
p'(-1) = f'(2)(-3) = (-2)(-3) = 6
\)