Answer
$x=0$, $x=4$ and $x=\frac{8}{7}$ are critical numbers of $F$.
Work Step by Step
How to find the critical numbers of a function $f$ according to definition
1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist.
2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not.
- If $c$ lies in $D_f$, $c$ is a critical number of $f$.
- If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$.
$$F(x)=x^{4/5}(x-4)^2$$
$$F(x)=\sqrt[5]{x^4}(x-4)^2$$
$D_F=R$
1) Now, we find $F'(x)$
$$F'(x)=\frac{4}{5}x^{-1/5}(x-4)^2+x^{4/5}2(x-4)$$
$$F'(x)=\frac{4}{5}x^{-1/5}(x-4)^2+2x^{4/5}(x-4)$$
$$F'(x)=\frac{4}{5}\sqrt[5]{\frac{1}{x}}(x-4)^2+2x^{4/5}(x-4)$$
We find $F'(x)=0$, $$F'(x)=0$$
$$\frac{4}{5}x^{-1/5}(x-4)^2+2x^{4/5}(x-4)=0$$
$$\frac{4(x-4)^2}{5x^{1/5}}+2x^{4/5}(x-4)=0$$
$$(x-4)[\frac{4(x-4)}{5x^{1/5}}+2x^{4/5}]=0$$
$$(x-4)[\frac{4(x-4)+10x}{5x^{1/5}}]=0$$
$$(x-4)[\frac{14x-16}{5x^{1/5}}]=0$$
$$x=4\hspace{0.5cm}or\hspace{0.5cm}x=\frac{8}{7}$$
Also, looking at $$F'(x)=\frac{4}{5}\sqrt[5]{\frac{1}{x}}(x-4)^2+2x^{4/5}(x-4)$$
reveals that for $x=0$, F'(x) would not exist.
2) Examine whether values of $x$ lie in $D_F$ or not.
We see that all of $x=0$, $x=4$ and $x=\frac{8}{7}\in R$, so they all lie in $D_F$.
We conclude that $x=0$, $x=4$ and $x=\frac{8}{7}$ are critical numbers of $F$.