## Calculus (3rd Edition)

$x=2,\ \ x=2/3$ $f(x)$ has a local maxima at $x= 2/3$ and has a local minima at $x= 2$
Given $$f(x)=x^{3}-4 x^{2}+4 x$$ Since \begin{align*} f'(x) &= 3x^2-8x+4\\ f''(x)&= 6x-8 \end{align*} Since $f(x)$ has a critical point when $f'(x)= 0$ \begin{align*} f'(x)&=0\\ 3x^2-8x+4&=0\\ (3x-2)(x-2)&= 0\end{align*} Then $f(x)$ has critical points at $x= 2,\ x=2/3$, and \begin{align*} f''(3/2)&= -4<0\\ f''(2)&=4>0 \end{align*} Then $f(x)$ has a local maxima at $x= 2/3$ and has a local minima at $x= 2$.