## Calculus (3rd Edition)

$x= 4/3$
Given $$y=x^{3}-4 x^{2}+4 x$$ Since \begin{align*} y'&=3x^{2}-8x+4 \\ y''&=6x-8 \end{align*} Then $f''(x)=0$ for $x=4/3$ and $$y''>0\ \ \text{ for } \ x>4/3\\ y''<0\ \ \text{ for } \ x<4/3$$ Hence, there exists an inflection point at $x= 4/3$