#### Answer

$f(x) $ is decreasing on $(-\infty,5) $ and increasing on $ ( 5,\infty) $
$f(x)$ has a local minimum at $x=5$.

#### Work Step by Step

Given $$y=x^{2}+(10-x)^{2} $$
Since
$$f'(x) =2x-2(10-x) $$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
2x-2(10-x)&=0\\
4x-20&=0
\end{align*}
Then $x= 5$ is a critical point. To find the interval where $f$ is increasing and decreasing, choose $x=0$ and $x= 6$
\begin{align*}
f'(0)&=-20<0\\
f'(6)&=4>0
\end{align*}
Hence, $f(x) $ is decreasing on $(-\infty,5) $ and increasing on $ ( 5,\infty) $. Hence, by using the first derivative test, $f(x)$ has a local minimum at $x=5$.