Answer
$y(4)=0$ is a local minimum and there are no local maxima; absolute minimum is at $y(4)=0$ and there is no absolute maximum.
concave down on $(-\infty,4)$ and $(4,\infty)$.
See graph.
Work Step by Step
Step 1. Given the function $y=\sqrt {|x-4|}=\begin{cases} \sqrt {x-4}\ x\geq4 \\ \sqrt {4-x}, \ x\lt4\end{cases}$, we have $y'=\begin{cases} \frac{1}{2}(x-4)^{-1/2},\ x\geq4 \\ -\frac{1}{2}(4-x)^{-1/2}, \ x\lt4\end{cases}$ and $y''=\begin{cases} -\frac{1}{4}(x-4)^{-3/2},\ x\geq4 \\ -\frac{1}{4}(4-x)^{-3/2}, \ x\lt4\end{cases}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=4$ as a critical points with $y(4)=0$
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(4)..(+)..$; we can identify $y(4)=0$ as a local minimum and there are no local maxima. The absolute minimum is at $y(4)=0$ and there is no absolute maximum.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We do not have possible infletion points within the domain as $y'$ is not continuous at $x=4$.
Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(-)..(4)..(-)..$,
Step 6. The function is concave down on $(-\infty,4)$ and $(4,\infty)$.
Step 7. See graph.
