#### Answer

(a) $x=2$ is a critical number of $f$ and at $x=2$ there is a horizontal tangent.
$x=2$ is the point where there is a local maximum of $f$.
(b) $x=6$ is a critical number of $f$ and at the point where $x=6$ there is a horizontal tangent.

#### Work Step by Step

(a) $f'(2)=0$ and $f''(2)=-5$
$f'(2)=0$ tells us that $x=2$ is a critical number of $f$. There will be a horizontal tangent there at $x=2$.
We do not know whether $f$ changes from increase to decrease or vice versa, since there is no information as to the signs of $f'$.
$f''(2)=-5\lt0$. According to the Second Derivative test, this shows that $x=2$ is the point where there is a local maximum of $f$.
(b) $f'(6)=0$ and $f''(6)=0$
Again, $f'(6)=0$ tells us that $x=6$ is a critical number of $f$. There will be a horizontal tangent there at $x=6$.
However, $f''(6)=0$ does not give out any information since, according to the Second Derivative Test, it is inconclusive. There must be further information to know more about the graph of $f$.