## Calculus (3rd Edition)

$$x= 0,\ x= 4$$
Given $$f(x)=\frac{x^{2}}{x^{2}-4 x+8}$$ Since \begin{align*} f'(x)&= \frac{\frac{d}{dx}\left(x^2\right)\left(x^2-4x+8\right)-\frac{d}{dx}\left(x^2-4x+8\right)x^2}{\left(x^2-4x+8\right)^2}\\ &=\frac{2x\left(x^2-4x+8\right)-\left(2x-4\right)x^2}{\left(x^2-4x+8\right)^2}\\ &=\frac{-4x^2+16x}{\left(x^2-4x+8\right)^2} \end{align*} Then to find the critical points \begin{align*} f'(x)&=0\\ \frac{-4x^2+16x}{\left(x^2-4x+8\right)^2}&=0\\ -4x^2+16x&=0\\ 4x(-x+4)&=0 \end{align*} Hence $f(x)$ has critical points at $x= 0,\ x= 4$