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For some reason this has me stuck. I can't seem to reverse what I know about derivates etc..

The graph of $f(t)$ appears below Integral Graph

If $g(x):=\int_{0}^{x}f(t)dt$, then find the following:

(a)$g(0)$

(b)g'(1)

(c)The interval where $g$ is convex

(d)The value of $x$ where $g$ takes its maximum on the interval $[0,8]$

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    This is a very good calculus problem!2011-12-01

1 Answers 1

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(a) $g(0)=\int_0^0f(t)dt=0$

(b) g'(1)=\frac{d}{dx}(\int^x_0f(t)dt)\Big|_{x=1}=f(1)=-19 (0r -20? Or 19.5? I need to have a magnifying glass to see the exact value from the graph. Anyway, it's $f(1)$.)

(c) Note that \displaystyle\frac{d^2g}{dx^2}(x)=\frac{d^2}{dx^2}(\int^x_0f(t)dt)=f'(x). Therefore, $\displaystyle\frac{d^2g}{dx^2}(x)>0$ if and only if f'(x). That is, the interval where $g$ is convex is the same as the interval where $f$ is increasing, that is $(1,6)$ from the graph. (Again a magnifying glass is needed)

(d) Again, $\displaystyle\frac{dg}{dx}(x)=f(x)$. That is, the critical point of $g$ correspond to the zero of $f$. Therefore, $2$ is critical point of $g$, and $g(2)=\int_0^2f(t)dt<0$ since $f<0$ on $(0,2)$. Finally, to find the absolute maximum of $g$ on $[0,8]$, we need to check the boundary point: $g(0)=0$ by part (a), and $g(8)>0$. Therefore, $g$ takes its maximum at $x=8$.