Answer
$y(-2\sqrt 2)=0, y(2)=4$ is a local maxima and $y(-2)=-4, y(2\sqrt 2)=0$ is a local minima.
absolute minimum is at $y(-2)=-4$ and the absolute maximum is at $y(2)=4$ .
$x=0$ is an inflection point.
concave up on $(-2\sqrt 2,0)$ and concave down on $(0,2\sqrt 2)$
See graph.
Work Step by Step
Step 1. Given the function $y=x\sqrt{8-x^2}=x(8-x^2)^{1/2}, -2\sqrt2\leq x\leq2\sqrt2$, we have $y'=(8-x^2)^{1/2}+\frac{1}{2}x(8-x^2)^{-1/2}(-2x)=\frac{8-x^2}{\sqrt{8-x^2}}-\frac{x^2}{\sqrt{8-x^2}}=\frac{8-2x^2}{\sqrt{8-x^2}}=2(4-x^2)(8-x^2)^{-1/2} $ and $y''=2(-2x)(8-x^2)^{-1/2}-(4-x^2)(8-x^2)^{-3/2}(-2x)=\frac{-4x(8-x^2)}{\sqrt{(8-x^2)^3}}+\frac{2x(4-x^2)}{\sqrt{(8-x^2)^3}}=\frac{2x^3-24x}{\sqrt{(8-x^2)^3}} $
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=-2\sqrt 2, -2, 0,2, 2\sqrt 2$ as critical points with $y(-2\sqrt 2)=0, y(-2)=-4, y(0)=0, y(2)=4, y(2\sqrt 2)=0$
Step 3. Examine the sign change of $y'$ across the critical points: $(-2\sqrt 2)..(-)..(-2)..(+)..(0)..(+)..(2)..(-)..(2\sqrt 2)$, we can identify $y(-2\sqrt 2)=0, y(2)=4$ as a local maxima and $y(-2)=-4, y(2\sqrt 2)=0$ as a local minima. The absolute minimum is at $y(-2)=-4$ and the absolute maximum is at $y(2)=4$ .
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=0$ as possible inflection points within the domain.
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 $..(+)..(0)..(-)..$,
Step 6. The function is concave up on $(-2\sqrt 2,0)$ and concave down on $(0,2\sqrt 2)$, this means that $x=0$ is an inflection point.
Step 7. See graph.
