Chapter 5.1, Problem 29E

To determine

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

Answer to Problem 29E

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

Explanation of Solution

Given information:

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

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,

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

For critical points equal the derivative of function to zero,

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

Now substitute in function,

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

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