Answer
Function $y$, at $x=-1$ and $x=3$, has the absolute and local minimum value of $0$. It also has the absolute and local maximum value of $2$ at $x=1$.
Work Step by Step
$$y=\sqrt{3+2x-x^2}=(3+2x-x^2)^{1/2}$$
$y$ is not defined where $(3+2x-x^2)\ge0$.
1) Find all the critical points:
Find the derivative of the function: $$y'=\frac{1}{2}(3+2x-x^2)^{-1/2}(3+2x-x^2)'$$ $$y'=\frac{1}{2}(3+2x-x^2)^{-1/2}(2-2x)$$ $$y'=\frac{2-2x}{2\sqrt{3+2x-x^2}}$$ $$y'=\frac{1-x}{\sqrt{3+2x-x^2}}$$
- For $y'=0$, $1-x=0$ or $x=1$.
- For $y'$ to be undefined, $\sqrt{3+2x-x^2}=0$, or $(3+2x-x^2)=0$.
We have $$3+2x-x^2=-(x^2-2x-3)=-\Big((x^2-2x+1)-4\Big)$$ $$=-\Big((x-1)^2-4\Big)$$
So $(3+2x-x^2)=0$ when $(x-1)^2=4$, or $(x-1)=\pm2$
For $(x-1)=2$, then $x=3$ and for $(x-1)=-2$, then $x=-1$.
All these points $x=\{-1,1,3\}$ are critical points here.
2) Evaluate function $y$ at the critical points:
- For $x=-1$: $$y=\sqrt{3+2\times(-1)-(-1)^2}=\sqrt{3-2-1}=0$$
- For $x=1$: $$y=\sqrt{3+2\times1-(1^2)}=\sqrt{3+2-1}=\sqrt4=2$$
- For $x=3$: $$y=\sqrt{3+2\times3-3^2}=\sqrt{3+6-9}=0$$
Now check the graph of function $y$.
We see that function $y$, at $x=-1$ and $x=3$, has the absolute and local minimum value of $0$. It also has the absolute and local maximum value of $2$ at $x=1$.
