maximum and minimum of $f(x)=\frac{x}{1+x^2}$

Question

Is there a way how to calculate maximum and minumum of

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

without derivative?

Answer 1

Let $$y=\frac{x}{1+x^2},$$ where $x$ is real. Then writing as a quadratic in $x$ gives $$yx^2-x+y=0,$$ whose discriminant $(-1)^2-4y^2$ must not be negative. Hence we find that $$1-4y^2\ge 0.$$ This tells us that $$|y|\le\frac12,$$ or in other words that $$-\frac12\le y\le \frac12.$$

These bounds will clearly be attained since then our quadratic will have real roots for $x.$ Thus, these are the extreme values of the function.

Answer 2

\begin{align*} \frac{x^{2} + 1}{2} \geq \sqrt{x^{2}} = |x| \Longrightarrow \left|\frac{x}{1+x^{2}}\right| \leq \frac{1}{2} \Longrightarrow -\frac{1}{2}\leq\frac{x}{1+x^{2}}\leq\frac{1}{2} \end{align*}

Answer 3

First, this function is odd. So it is enough to find its bounds on $\mathbf R_{\ge 0}$.

You can rewrite this function, for $x\ne 0$, as $$\frac x{1+x^2}= \frac 1{x+\dfrac 1x}.$$

Now it is standard that, for $x>0$, $x+\dfrac1x \ge 2$ (it is a direct consequence of the AGM inequality) and the minimum is attained when $x=\frac 1x$, i.e. $x=1$.

Therefore, if $x\ge 0$, we have $$0\le\frac x{1+x^2} \le \frac12,$$By symmetry, if $x\le 0$,we have $$-\frac12\le \frac x{1+x^2} \le 0,$$ from which you obtain instantly the bounds, attained at $1$ and $-1$.

Answer 4

For $x>0$,

$$f(x)=\frac{x}{1+x^2}=\frac1{(\sqrt x-\frac1{\sqrt x} )^2 +2 }\le \frac12 $$

Given that $f(x)$ is odd, we have

$$-\frac12 \le f(x)\le \frac12 $$