## Calculus: Early Transcendentals 8th Edition

$x=\sqrt e$ is the critical number of $f$.
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 $f$.