Chapter 4.1, Problem 29E

To determine

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

Answer to Problem 29E

The extreme value of the function is $-0.5$ and minimum at $x=-1$ .

The extreme value of the function is $0.5$ and maximum at $x=1$ .

Explanation of Solution

Given information :

  $y=\frac{x}{x^2+1}$

Formula needed :

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

The domain of the function is $ℝ$ .

Differentiate $y=\frac{x}{x^2+1}$ with respect to $x$ .

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

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

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

The critical points are $x=-1,1$ .

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

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

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

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

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 $y$ increases.

Thus, minimum occur at $x=-1$ .

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

Thus, maximum occur at $x=1$ .