Help finding critical numbers. Find the critical numbers of the function:
$$f(x)=x^\frac{4}{5}(x-4)^2$$
Help would be so greatly appreciated :]
 A: $(1)$ find $f'(x)$:


*

*Use the product rule to find $f'(x)$, 

*or else expand $f(x)$: $$f(x) = x^{4/5}(x^2 - 8 x + 16) = x^{14/5} -
   x^{9/5} + 16x^{4/5}$$ and find $f'(x)$.


$(2)$ Then set $f'(x) = 0$ and solve for x.
$(3)$ Check for values for which $f(x)$ is not defined (there are none).
$(4)$ Check for values for which $f'(x)$ is not defined. There is one such value $x'$. Examine what happens as you take the limit as $x \to x'$ from both the right and from the left of $x'$.
ADDED: 
As per comments below, your derivative is correct. Expressing it with a common denominator and setting equal to zero gives us:
$$f'(x) = \dfrac{10x(x-4) + 4(x-4)^2}{5x^{1/5}}$$
So we need to check what happens at $$x = 0$$ as described above, when $f'(x)$ is undefined. 
Then, the derivative will be zero (with two "zeros") when 
$$10x(x-4) + 4(x-4)^2 = 0$$
Factoring and simplifying gives us $$2(x-4)(5x + 4x - 16) = 2(x-4)(9x-16) = 0$$ 
which is has roots (values at which $f'(x) = 0$) when $$x = 4, \quad x = \dfrac{16}{9}$$
