Answer
$x=0$ and $x=\frac{2}{3}$ are critical numbers of $f$.
Work Step by Step
How to find the critical numbers of a function $f$ according to the 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 $f$.
$$f(x)=x^2e^{-3x}$$
$D_f=R$
*First, we find $f'(x)$
$$f'(x)=2xe^{-3x}+x^2e^{-3x}(-3x)'$$
$$f'(x)=2xe^{-3x}+x^2e^{-3x}(-3)$$
$$f'(x)=2xe^{-3x}-3x^2e^{-3x}$$
$$f'(x)=xe^{-3x}(2-3x).$$
We find $f'(x)=0$, $$f'(x)=0$$
$$x=0\hspace{0.5cm}or\hspace{0.5cm}e^{-3x}=0\hspace{0.5cm}or\hspace{0.5cm}(2-3x)=0$$
- For $e^{-3x}=0,$
this equation has no satisfying values of $x$ since $e^{-3x}\gt0$ for all $x$.
- For $2-3x=0$ or $x=\frac{2}{3,}$
this value belongs to $D_f$, so it is a critical number of $f$
- For $x=0,$
this value belongs to $D_f$, so it is a critical number of $f$.
We conclude that $x=0$ and $x=\frac{2}{3}$ are critical numbers of $f$.
