Answer
$x= 4/3$
Work Step by Step
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 $x= 4/3$
