Chapter 5.1, Problem 27E

To determine

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

Answer to Problem 27E

the minimum value is $0$ at $x=-1,3$ and maximum value is $\sqrt{3}$ at $x=0$ .

Explanation of Solution

Given information:

The function is $y=\sqrt{3+2x-x^2}$

Calculation: the function is defined for each value of x,

  $\begin{array}{l}3+2x-x^2\ge 0\\ x^2-3x+x-3\le 0\text{ [simplify]}\\ \left(x-3\right)\left(x+1\right)\le 0\text{ [factorise]}\\ -1\le x\le 3\end{array}$

The function domain is $\left[-1,3\right]$ , so the domain of function has end points, so the extreme points occur at end points,

To find the extreme values of function, first find its derivative,

  $\begin{array}{l}y=\sqrt{3+2x-x^2}\\ y\text{'}=\frac{x-2x}{2\sqrt{3+2x-x^2}}\text{ [differentiate and simplify]}\end{array}$

For critical points equal the derivative of function to zero,

  $\begin{array}{l}\frac{x-2x}{2\sqrt{3+2x-x^2}}=0\\ x=0\text{ [simplify]}\end{array}$

Now substitute in function,

  $\begin{array}{l}y=\sqrt{3+2x-x^2}\\ y=\sqrt{3}\text{ [substitute }x=0\text{ ]}\\ y=0\text{ [substitute }x=-1]\\ y=0\text{ [substitute }x=3]\end{array}$

Thus, the minimum value is $0$ at $x=-1,3$ and maximum value is $\sqrt{3}$ at $x=0$ .