Chapter 4.1, Problem 27E

To determine

To find : extreme values of the function and where they occur.

Answer to Problem 27E

The extreme value of the function is $2$ and maximum at $x=1$ .

The extreme value of the function is $0$ and minimum at $x=-1$ and, $x=3$ .

Explanation of Solution

Given information :

  $y=\sqrt{3+2x-x^2}$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The domain of the function is $\left[-1,3\right]$ .

Differentiate $y=\sqrt{3+2x-x^2}$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{3+2x-x^2}\right)\\ =\frac{1}{2\sqrt{3+2x-x^2}}\cdot \frac{d}{dx}\left(3+2x-x^2\right)\\ =\frac{1}{2\sqrt{3+2x-x^2}}\cdot \left(2-2x\right)\\ =\frac{1-x}{\sqrt{3+2x-x^2}}\end{array}$

Take $\frac{dy}{dx}=0$ for critical points,

  $\begin{array}{c}\frac{1-x}{\sqrt{3+2x-x^2}}=0\\ 1-x=0\left[\begin{array}{l}3+2x-x^2\ne 0\\ x\ne -1,3\end{array}\right]\\ x=1\end{array}$

The critical point is $x=-1,1,3$ .

Now plug point $x=-1$ in $y=\sqrt{3+2x-x^2}$ .

  $\begin{array}{c}y=\sqrt{3+2\left(-1\right)-{\left(-1\right)}^2}\\ =\sqrt{3-2-1}\\ =0\end{array}$

Now plug point $x=1$ in $y=\sqrt{3+2x-x^2}$ .

  $\begin{array}{c}y=\sqrt{3+2\left(1\right)-{\left(1\right)}^2}\\ =\sqrt{3+2-1}\\ =2\end{array}$

Now plug point $x=3$ in $y=\sqrt{3+2x-x^2}$ .

  $\begin{array}{c}y=\sqrt{3+2\left(3\right)-{\left(3\right)}^2}\\ =\sqrt{3+6-9}\\ =0\end{array}$

Examine whether the function is either maxima or minima. Check the surrounding values of critical point.

As $x$ moves away from $1$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=1$ .

At end points, the value of function is $0$ .

Therefore, the minimum occurs at $x=-1$ and, $x=3$ .