# Calc2.52

Find the local minimums and maximums of the function $f(x)={\sqrt {x}}$.
$f'(x)={\frac {1}{2{\sqrt {x}}}}\,$
This function is never equal to 0 as the numerator is never equal to 0. It is undefined at $x=0$. This is our only critical point. However, the square root function is only defined for nonnegative real numbers so $f(x)$ is not defined to the left of $x=0$. Thus, it can not have a local minimum or maximum at $x=0$. $x=0$ does give us the lowest point on the entire square root function as it starts at $0$ and increases thereafter.