Question

In Exercises \(19-30\) , find the extreme values of the function and where they occur. $$y=\frac{x}{x^{2}+1}$$

Short answer

The function \(y=\frac{x}{x^{2}+1}\) has no extreme values.

Step-by-step solution

Compute the derivative

The first step is to calculate the derivative of \(y=\frac{x}{x^{2}+1}\) using the quotient rule, which states \((\frac{u}{v})'=\frac{u'v-uv'}{v^{2}}\), where u=x, v=\(x^{2}+1\): \[y'=\frac{(1)(x^{2}+1)-(2x)(x)}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{(x^{2}+1)^{2}}.\]

Find the critical points

The next step is to set the derivative equal to zero to find the critical points. This will find any local maximum or minimum points, as well as possible points of inflection: \[\frac{1-x^{2}}{(x^{2}+1)^{2}}=0.\] Solve for x, we get x = 1 and x = -1.

Use the second derivative test

Next, we need to find the second derivative \[y''\] of the function and substitute the critical points into the second derivative. By calculating the answer we can determine the nature of these points. If \(y''(x)>0\) then x is a local minimum, and if \(y''(x)<0\) then x is a local maximum. By calculating, we get \(y''=-2(x^{2}-1)/(x^{2}+1)^{3}\). So \(y''(1)=y''(-1)=0\). Since the second derivative evaluated at the critical points is neither positive nor negative, these two points are inflection points, not extreme points.

Finding the extreme values

Since there are no local maximum or minimum, the function \(y=\frac{x}{x^{2}+1}\) has no extreme values.