Answer
- At $x=-2$, $y$ has a local maximum value of $0$.
- At $x=2$, $y$ has a local minimum value of $0$.
- At $x=-\sqrt2$, $y$ has a local and absolute minimum value of $-2$.
- At $x=\sqrt2$, $y$ has a local and absolute maximum value of $2$.
The critical points are $(-2,0)$, $(-\sqrt2,-2)$, $(\sqrt2,2)$ and $(2,0)$. The endpoints are $(-2,0)$ and $(2,0)$.
Work Step by Step
$$y=x\sqrt{4-x^2}$$
- Domain: $[-2,2]$. So the endpoints are at $x=2$ and $x=-2$.
1) Find all the critical points of the function:
- Find $y'$: $$y'=(x)'\sqrt{4-x^2}+x(\sqrt{4-x^2})'$$ $$y'=\sqrt{4-x^2}+\frac{x(4-x^2)'}{2\sqrt{4-x^2}}=\sqrt{4-x^2}+\frac{x(-2x)'}{2\sqrt{4-x^2}}$$ $$y'=\sqrt{4-x^2}-\frac{2x^2}{2\sqrt{4-x^2}}$$ $$y'=\frac{2(4-x^2)-2x^2}{2\sqrt{4-x^2}}=\frac{8-2x^2-2x^2}{2\sqrt{4-x^2}}$$ $$y'=\frac{8-4x^2}{2\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}$$
- We have $y'=0$ when $$4-2x^2=0$$ $$x^2=2$$ $$x=\pm\sqrt2$$.
- $y$ is undefined when $$\sqrt{4-x^2}=0$$ $$4-x^2=0$$ $$x^2=4$$ $$x=\pm2$$
So the critical points of function $y$ are at $x=\pm\sqrt2$ and $x=\pm2$. $x=\pm2$ are also the places where the endpoints lie.
2) Evaluate $y$ at the endpoints and critical points:
- For $x=-2$: $$y=(-2)\sqrt{4-(-2)^2}=(-2)\sqrt{4-4}=0$$
- For $x=-\sqrt2$: $$y=(-\sqrt2)\sqrt{4-(-\sqrt2)^2}=(-\sqrt2)\sqrt{4-2}=(-\sqrt2)\sqrt2=-2$$
- For $x=\sqrt2$: $$y=(\sqrt2)\sqrt{4-(\sqrt2)^2}=(\sqrt2)\sqrt{4-2}=(\sqrt2)\sqrt2=2$$
- For $x=2$: $$y=2\sqrt{4-2^2}=2\sqrt{4-4}=0$$
So the critical points are $(-2,0)$, $(-\sqrt2,-2)$, $(\sqrt2,2)$ and $(2,0)$. The endpoints are $(-2,0)$ and $(2,0)$.
Now take a look at the graph of $y$.
From the graph, we see that
- At $x=-2$, $y$ has a local maximum value of $0$.
- At $x=2$, $y$ has a local minimum value of $0$.
- At $x=-\sqrt2$, $y$ has a local and absolute minimum value of $2$.
- At $x=\sqrt2$, $y$ has a local and absolute maximum value of $2$.
