## University Calculus: Early Transcendentals (3rd Edition)

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 $x=1$.
$$y=\sqrt{3+2x-x^2}=(3+2x-x^2)^{1/2}$$ $y$ 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 $x=1$.