Answer
(a) $G$ is decreasing on the intervals $(-\infty,0)\cup (1, \infty)$
$G$ is increasing on the interval $(0,1)$
(b) The local minimum is $G(0) = 0$
The local maximum is $G(1) = 3$
(c) The graph is concave down on these intervals: $(-\frac{1}{2}, 0)\cup (0, \infty)$
The graph is concave up on the interval: $(-\infty, -\frac{1}{2})$
The point of inflection is $(-0.5, 3.78)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $G(x) = 5x^{2/3}-2x^{5/3}$
We can find the points where $G'(x) = 0$:
$G'(x) = (\frac{2}{3})(5x^{-1/3})-(\frac{5}{3})(2x^{2/3})$
$G'(x) = \frac{10}{3x^{1/3}}-\frac{10}{3}x^{2/3} = 0$
$\frac{10}{3x^{1/3}}=\frac{10}{3}x^{2/3}$
$\frac{1}{x^{1/3}}=x^{2/3}$
$x = 1$
Also note that $G'(x)$ is undefined at $x=0$
When $-\infty \lt x \lt 0~~$ or $x\gt 1~~$ then $G'(x) \lt 0$
$G$ is decreasing on the intervals $(-\infty,0)\cup (1, \infty)$
When $0 \lt x \lt 1~~$ then $G'(x) \gt 0$
$G$ is increasing on the interval $(0,1)$
(b) $G(0) = 5(0)^{2/3}-2(0)^{5/3} = 0$
$G(1) = 5(1)^{2/3}-2(1)^{5/3} = 3$
The local minimum is $G(0) = 0$
The local maximum is $G(1) = 3$
(c) We can find the points where $G''(x) = 0$:
$G''(x) = -\frac{10}{9x^{4/3}}-\frac{20}{9}x^{-1/3}$
$G''(x) = -\frac{10}{9x^{4/3}}-\frac{20}{9x^{1/3}} = 0$
$-\frac{10}{9x^{4/3}}= \frac{20}{9x^{1/3}}$
$x = -\frac{1}{2}$
Also note that $G''(x)$ is undefined at $x=0$
The graph is concave down when $G''(x) \lt 0$
The graph is concave down on these intervals: $(-\frac{1}{2}, 0)\cup (0, \infty)$
The graph is concave up when $G''(x) \gt 0$
The graph is concave up on the interval: $(-\infty, -\frac{1}{2})$
$G(-\frac{1}{2}) = 3.78$
The point of inflection is $(-0.5, 3.78)$
(d) We can see a sketch of the graph below.