Chapter 5.1, Problem 30E

To determine

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

Answer to Problem 30E

the minimum value is $-\frac{1}{2}$ at $x=-2$ and maximum value is $\frac{1}{2}$ at $x=0$ .

Explanation of Solution

Given information:

The function is $y=\frac{x+1}{x^2+2x+2}$

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

The function domain is $R$ , so the domain of function has no end points, so the extreme points occur at critical points,

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

  $\begin{array}{l}y=\frac{x+1}{x^2+2x+2}\\ y\text{'}=\frac{-x^2-2x}{{\left(x^2+2x+2\right)}^2}\text{ [differentiate and simplify]}\end{array}$

For critical points equal the derivative of function to zero,

  $\begin{array}{l}\frac{-x^2-2x}{{\left(x^2+2x+2\right)}^2}=0\\ x=-2,0\text{ [simplify]}\end{array}$

Now substitute in function,

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

Thus, the minimum value is $-\frac{1}{2}$ at $x=-2$ and maximum value is $\frac{1}{2}$ at $x=0$ .