Answer
$f(x)$ is increasing on $( 0.5,2)\cup (2,\infty)$ and decreasing on $(-\infty, 0.5)$. By using the First Derivative Test, we see that $x= 0.5$ is a local minimum.
Work Step by Step
Given $$ y=x(x-2)^{3}$$
Then
$$ f'(x)= 3x(x-2)^{2}+(x-2)^{3}= (4 x-2)(x-2)^{2}$$
Then $f'(x)=0$ for $x=0.5$ and $x=2$. Choose $x=0$, $x=1$ and $x= 3$:
\begin{align*}
f'(0)&<0\\
f'(1)&>0\\
f'(3)&>0
\end{align*}
Hence, the function is increasing on $( 0.5,2)\cup (2,\infty)$ and decreasing on $(-\infty, 0.5)$. By using the First Derivative Test, we find that $x= 0.5$ is a local minimum.
