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

Question

Is there a way how to calculate maximum and minumum of

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

without taking the derivative of it?

Answer 1

$$0 \le x^2< \infty$$

$$1 \le 1+x^2< \infty$$

$$1 \ge \frac {1}{1+x^2}>0$$

Answer 2

Let's think about this:

When is a fraction $\frac{1}{A}$ the maximum? --> When A is the smallest.

When is a fraction $\frac{1}{A}$ the minimum? --> When A is the largest.

The minimal value for $1 + x^2$ is $1$ because $x^2$ is non-negative. So the minimum of $x^2$ is $0$ which leads to $f(x) = \frac{1}{1 + 0} = 1$. Therefore at $x = 0$ is the maximum of $\frac{1}{1+x^2}$.

However you can make $1+x^2$ as large as you want by enlarging $x$. So there is no largest number for $1 + x^2$ and therefore the minimum is in the limit approaching $0$ at $\pm \infty$.

Answer 3

Maximum of your function corresponds to the minimum of $1 + x^2$ and similarly for the minimum.

So you know that $x^2+1$ is a parabola and we know that min/max of a parabola $ax^2 + bx + c$ is at $\frac{-b}{2a}$. It can be shown without a derivative by representing it in the form $a(x-\lambda)^2 + \beta$.

Answer 4

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

Note that $0\le x^2<\infty$

Adding $1$ across the board,

$1\le 1+x^2 < \infty$

Taking the reciprocal,

$1 \ge \frac{1}{1+x^2} > 0$

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For $f(x)=\frac{x}{1+x^2}$ where $x>0$

Divide both numerator and denominator by $x$,

So, $f(x)=\frac{1}{\frac{1}{x}+x}$,

Note that the denominator can be rewritten as

$(\sqrt{x}+\frac{1}{\sqrt{x}}+2)-2=(\sqrt{x}+\frac{1}{\sqrt{x}})^2-2$.

Clearly, $(\sqrt{x}+\frac{1}{\sqrt{x}})^2-2 \ge -2$ or $(\sqrt{x}+\frac{1}{\sqrt{x}})^2 \ge 0$ or $x+\frac{1}{x}+2 \ge 0$

This means, $x+\frac{1}{x} \ge 2$ or $x+\frac{1}{x} \le -2$

The equality holds true (for max. and min.) leading us to:

• Max is $1$ by setting $x=1$.

• Min is $-$ by setting $x=-1$.

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While answering this problem, the original post has been edited where this part was removed, hope you make your question clear as much as possible before posting, thanks, hope my answer helps.