Question
Let f be a continuous function defined on the closed interval −4 ≤ x ≤ 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)• 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
Ans:
(a)
G'(x) = f(x)
G”(x) = f'(x)
On (-4, -2) and (2, 6), G(x) is concave up because f(x) (which is equal to G'(x)) has a positive slope / is increasing.
(b)
p'(x) = G'(x) f(x) + f'(x) G(x)
p'(3) = G'(3) f(3) + f'(3) G(3)
G'(x) = f(x) \(G(3)=\int_{0}^{3}f(t)dt=-\frac{7}{2}\)
\(p'(3)=(-3)(-3)+(1)\left (- \frac{7}{2} \right )\)
(c)
\(\lim_{x\rightarrow 2}G(x)=\lim_{x\rightarrow 2}(x^{2}-2x)=0\) Must be I’hopital’s rule
\(\int_{0}^{2}f(t)dt=0\)
\(\lim_{x\rightarrow 2}\frac{G'(x)}{2x-2}=\frac{f(2)}{4-2}=\frac{-4}{2}\)
(d)
\(\int_{0}^{2}f(t)dt\) \(\int_{0}^{-4}f(t)dt= -(3+9+3+1)\)
= -16
AROC of \(G = \frac{G(2)-G(-4)}{2-(-4)}=\frac{0-(16)}{2+4}=\frac{16}{6}=\frac{8}{3}\)
The meanvalue theorem does guarantee a value of -4<c<2, for which G'(c) is equal to average rate of chang. This is because G'(x) = f(t) and x = t exists for all values, -4<x=+t<2, meaning that G(x) is continuous on the closed interval and differentiable on the open interval.