Answer
(a) $x=3$
(b) Increasing $(3,+\infty)$ and decreasing $(-\infty, 3)$
(c) Concave down $(-\infty,-1),(7,+\infty)$ and concave up (-1,7)
Work Step by Step
(a) The fundamental theorem of calculus part II gives: $F^{\prime}(x)=\frac{-3+x}{7+x^{2}}$
The only critical value is $x=3$, so the minimum value should be at $x=3$
(b) $F^{\prime}(x)=\frac{-3+x}{7+x^{2}}$ is negative on $(-\infty, 3)$, so the function is decreasing on $(-\infty, 3)$
$F^{\prime}(x)=\frac{-3+x}{7+x^{2}}$ is positive on $(-\infty, 3)$, so the function is increasing
on $(3,+\infty)$
(c) $F^{\prime \prime}(x)=\frac{6 x+7-x^{2}}{\left(x^{2}+7\right)^{2}}$, which has roots $x=7$, $x=-1$
The function is concave down if the second derivative is negative, so it is concave down on $(7,+\infty)$, $(-\infty,-1).$ The function is concave up if the second derivative is positive, which occurs for $(-1,7) $
