Answer
a. True.
b. True.
c. True.
d. False.
e. True.
f. False.
g. True.
Work Step by Step
Given $f'(t)\gt0, f(1)=0$ and $g(x)=\int_0^x f(t)dt$, we have:
a. True, because $g'(x)=f(x)$ which means that $g(x)$ is a differentiable function of x.
b. True. Based on the Fundamental Theorem Calculus, $g(x)=\int_0^x f(t)dt$ is a continuous function of x.
c. True, because $g'(1)=f(1)=0$, which means that $g(x)$ has a horizontal tangent at $x=1$.
d. False. Test signs of $g'(x)$ across $x=1$; we have $..(-)..(0)..(+)..$ and the curve is concave up and does not have a maximum in this region.
e. True. Test signs of $g'(x)$ across $x=1$; we have $..(-)..(0)..(+)..$ and the curve is concave up and has a minimum at $x=1$.
f. False. Test signs of $g''(x)=f'(x)$ across $x=1$; we have $..(+)..(f '(1))..(+)..$ and there is no concavity change.
g. True, because $\frac{dg}{dx}=g'(1)=f(1)=0, g'(x)=f(x)\gt0$ the function $\frac{dg}{dx}$ crosses the x-axis from below to above at $x=1$.
