Answer
$ a.\quad$Minima at $(-2,0)$ and $(2,0)$. Maximum at $(0,2).$
$ b.\quad$Minimum at $(-2,0)$. Maximum at $(0,2).$
$ c.\quad$Maximum at $(0,2).$
$ d.\quad$Maximum at $(1,\sqrt{3}).$
Work Step by Step
$f(x)=(4-x^{2})^{1/2}$
$ f'(x)=(4-x^{2})^{-1/2}\cdot(-2x)\qquad$ ... chain rule
$=\displaystyle \frac{-2x}{\sqrt{4-x^{2}}}$
Critical point: $x=0$
$a.\quad \left[\begin{array}{lll}
\text{left endp.} & \text{critical p.} & \text{right endp.}\\
x=-2 & x=0 & x=2\\
f(-2)=0 & 2 & 0\\
\text{abs. min} & \text{abs. max} & \text{abs. min}
\end{array}\right]$
$b.\quad \left[\begin{array}{lll}
\text{left endp.} & \text{critical p.} & \text{right endp.}\\
x=-2 & x=0 & \text{- none -}\\
f(-2)=0 & 2 & \\
\text{abs. min} & \text{abs. max} &
\end{array}\right]$
$c.\quad \left[\begin{array}{lll}
\text{left endp.} & \text{critical p.} & \text{right endp.}\\
\text{- none -} & x=0 & \text{- none -}\\
& 2 & \\
& \text{abs. max} &
\end{array}\right]$
$d.\quad \left[\begin{array}{lll}
\text{left endp.} & \text{critical p.} & \text{right endp.}\\
x=1 & \text{- none -} & \text{- none -}\\
f(1)=\sqrt{3} & & \\
& &
\end{array}\right]$
