Answer
$x=\sqrt e$ is the critical number 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 $f$.
$$f(x)=x^{-2}\ln x$$
$\ln x$ is only defined for $x\gt0$. Therefore, $$D_f=(0,+\infty)$$
*First, we find $f'(x).$
$$f'(x)=-2x^{-3}\ln x+x^{-2}\frac{1}{x}$$
$$f'(x)=-2x^{-3}\ln x+x^{-3}$$
$$f'(x)=x^{-3}(-2\ln x+1)$$
$$f'(x)=\frac{1}{x^3}(-2\ln x+1)$$
We find $f'(x)=0$, $$f'(x)=0$$
$$-2\ln x+1=0$$ (since $\frac{1}{x^3}$ cannot equal $0$).
$$\ln x=\frac{1}{2}$$
$$x=e^{1/2}=\sqrt e$$
This value belongs to $D_f$, so it is a critical number of $f$.
Also, looking at $$f'(x)=\frac{1}{x^3}(-2\ln x+1)$$, we see for $x=0$, $f'(x)$ does not exist. However, $x=0$ does not belong to $D_f$, so it is not a critical number of $f$.
We conclude that $x=\sqrt e$ is the critical number of $f$.
