Critical point $c=3$ is a point of local maximum.
Work Step by Step
$f'(x)=6-2x$ is defined everywhere. $f'(x)=0\implies x=3$ is the critical point of $f(x)$. $f'(2.99)=6-(2\times2.99)=positive$ $f'(3.01)=6-(2\times3.01)=negative$ The sign change of $f'(x)$ at the critical point is positive to negative. Therefore, the critical point is a point of local maximum.