Answer
(a) $F$ is decreasing on the interval $(4,6)$
$F$ is increasing on the interval $(-\infty, 4)$
(b) The local maximum is $F(4) = 4\sqrt{2}$
(c) The graph is concave down on this interval: $(-\infty, 6)$
There are no points of inflection.
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $F(x) = x~\sqrt{6-x}$
Note that this function is defined on the interval $(-\infty, 6]$
We can find the points where $F'(x) = 0$:
$F'(x) = \sqrt{6-x}-\frac{x}{2\sqrt{6-x}}$
$F'(x) = \frac{2(6-x)}{2\sqrt{6-x}}-\frac{x}{2\sqrt{6-x}}$
$F'(x) = \frac{12-3x}{2\sqrt{6-x}} = 0$
$12-3x = 0$
$x = 4$
When $4 \lt x \lt 6~~$ then $F'(x) \lt 0$
$F$ is decreasing on the interval $(4,6)$
When $x \lt 4~~$ then $F'(x) \gt 0$
$F$ is increasing on the interval $(-\infty, 4)$
(b)$F(4) = (4)~\sqrt{6-4} = 4\sqrt{2}$
The local maximum is $F(4) = 4\sqrt{2}$
(c) We can find the points where $F''(x) = 0$:
$F''(x) = \frac{(-3)(2\sqrt{6-x})-(12-3x)(\frac{-1}{\sqrt{6-x}})}{4(6-x)}$
$F''(x) = \frac{\frac{(-6)(6-x)}{\sqrt{6-x}}+\frac{12-3x}{\sqrt{6-x}})}{4(6-x)}$
$F''(x) = \frac{\frac{-36+6x}{\sqrt{6-x}}+\frac{12-3x}{\sqrt{6-x}})}{4(6-x)}$
$F''(x) = \frac{-36+6x+12-3x}{4(6-x)^{3/2}}$
$F''(x) = \frac{3x-24}{4(6-x)^{3/2}} = 0$
$3x-24 = 0$
$x = 8$
Note that the function $F(x)$ is not defined at $x=8$
The graph is concave down when $F''(x) \lt 0$
The graph is concave down on this interval: $(-\infty, 6)$
The graph is concave up when $F''(x) \gt 0$
The graph is not concave up on any interval.
There are no points of inflection.
(d) We can see a sketch of the graph below.
