Question:
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
Ans:
\(g(x)= \int_{-3}^{3}f(t)dt\)
\(= \frac{1}{2}(5)(4)-\frac{1}{2}(1)(2)\)
= 10 – 1 = 9
(b)
g'(x) > 0 ⇔ f(x) > 0
g”(x) < 0 ⇔ f'(x) < 0
(-5, -3), (0, 2)
(c)
\(h(x)=\frac{g(x)}{5x}\)
\(h'(x)= \frac{(5x)(g'(x))-g(x).5}{25x^{2}}\)
\(h'(3)= \frac{(15)(f(3))-g(3).5}{9.25}=\frac{(15)(-2)-(9)(5)}{9.25}\)
(d)
p(x) = f(x2 – x)
p'(x) = f'(x2-x) . (2x-1)
p'(-1) = f'(1+1) . (-2-1) = f'(2).-3
= (-2) (-3)
= 6
\({f}'(2)=\frac{-4-4}{4-0}=\frac{-8}{4}=-2\)