Question: I just wanted to ask, why does the following function:

$$f(x)=x^{1/3}(x+3)^{2/3}$$

Have 3 crtical points $0,-1,-3$, because its first derivative is:

$$f'(x)=\frac{x+1}{x^{2/3}(x+3)^{1/3}}$$ 

$$f'(x)=\frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$

Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$

Answer: 
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
**Hint** *critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.*
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.