Question

Find all roots, local maxima and minima, and inflection points of each function f . In addition, determine whether any local extrema are also global extrema on the domain of f .

$f\left(x\right)=\frac{x}{x^2+1}$

Short answer

roots of the function are x=0

local maxima at x=1

local minima at x=-1

inflection point$x=0,\pm \sqrt{3}$

Step-by-step solution

Step1. Given Information

Consider a function

$f\left(x\right)=\frac{x}{x^2+1}$

$\text{For finding roots of the function;}$

$\begin{array}{r}f\left(x\right)=0\\ \frac{x}{x^2+1}=0\\ x=0\left[\text{multiplying both sides by}\left(x^2+1\right)\right]\end{array}$

The first derivative of the function is given below:

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{d}{dx}\left(\frac{x}{x^2+1}\right)\\ =\frac{\left(x^2+1\right)\cdot 1-x\left(2x\right)}{{\left(x^2+1\right)}^2}\\ =\frac{x^2+1-2x^2}{{\left(x^2+1\right)}^2}\\ =\frac{1-x^2}{{\left(x^2+1\right)}^2}\end{array}$

Step2. Critical Points

$\text{This derivative is always defined and continuous, so the critical points of}f\text{are just the places}$

where$f\text{'}\left(x\right)=0;$that is

$\begin{array}{r}\frac{1-x^2}{{\left(x^2+1\right)}^2}=0\\ 1-x^2=0\left[\text{multplying both sides by}{\left(x^2+1\right)}^2\right]\\ x^2=1\\ x=\pm 1\end{array}$

Step 3. Maxima and Minima

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

Therefore, the local maxima at x=1 and the local minima at x=-1.

$\begin{array}{r}{f}^{\prime \prime }\left(x\right)=\frac{d}{dx}\left[\frac{1-x^2}{{\left(x^2+1\right)}^2}\right]\\ =\frac{{\left(x^2+1\right)}^2(-2x)-\left(1-x^2\right)2\left(x^2+1\right)\cdot 2x}{{\left(x^2+1\right)}^4}\\ =\frac{\left(x^2+1\right)\left[\left(x^2+1\right)(-2x)-\left(1-x^2\right)4x\right]}{{\left(x^2+1\right)}^4}\\ =\frac{-2x^3-2x-4x+4x^3}{{\left(x^2+1\right)}^3}\\ =\frac{2x^3-6x}{{\left(x^2+1\right)}^3}\end{array}$

Step4. Inflection Point

For inflection point

$\begin{array}{r}{f}^{\prime \prime }\left(x\right)=0\\ \frac{2x^3-6x}{{\left(x^2+1\right)}^3}=0\\ 2x^3-6x=0\\ \left.2x\left(x^2-3\right)=0\text{[dividing both sides by}2\right]\\ x=0;x^2-3=0\\ x=0;x=\pm \sqrt{3}\end{array}$

Step5. Answer

$\text{Therefore, the roots of the function are}x=0\text{; the local maximum at}x=1\text{; and the local}$

$\text{minimum at}x=-1\text{. The inflection point of the function is}x=0,\pm \sqrt{3}\text{. The local maximum and}$

$\text{local minimum are also called as global maximum and global minimum.}$