Answer
a) Increasing on $(-\infty,4)$
decreasing on $(4,6)$
b) Local maximum value:$F(4) = 4\sqrt 2$;
no local minimum value
c) Concave downward on $(-\infty,6)$
No inflection point
d) See graph
Work Step by Step
a)
$F(x)$ = $x\sqrt {6-x}$
$F'(x)$ = $x(\frac{1}{2})(6-x)^{-\frac{1}{2}}+(6-x)^{\frac{1}{2}}$ = $\frac{-3x+12}{2\sqrt {6-x}}$
$F'(x)$ $\gt$ $0$
$-3x+12$ $\gt$ $0\Rightarrow x$ $\lt$ $4$
$F'(x)$ $\lt$ $0\Rightarrow 4$ $\lt$ $x$ $\lt$ $6$
$F$ is increasing on $(-\infty,4)$.
$F$ is decreasing on $(4,6)$.
b)
$F$ changes from increasing to decreasing at $x$ =$4$
$F(4) = 4\sqrt 2$ is a local maximum value
There is no local minimum value
c)
$F'(x)$ = $\frac{3}{2}(x-4)(6-x)^{-\frac{1}{2}}$
$F''(x)$ = $-\frac{3}{2}[(x-4)\frac{1}{2}(6-x)^{-\frac{3}{2}}+(6-x)^{-\frac{1}{2}}]$ = $\frac{3(x-8)}{4(6-x)^{\frac{3}{2}}}$
$F''(x)$ $\lt$ $0$ on $(-\infty,6)$
$F$ is concave downward on $(-\infty,6)$
There is no inflection point.
