## Calculus (3rd Edition)

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