# Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.

by Arc Neoepi   Last Updated July 17, 2017 10:20 AM

Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .

What I tried:

Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.

Secondly, decided to use differentiation

$y=\frac{2x}{x^2 +1}$

$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$

For stationary points,

$\frac {dy}{dx} = 0$

$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$

$x=-1$ or $x=1$

When $x=-1,y=-1$

When $x=1,y=1$

Therefore this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.

^I wonder if this is the correct method or did I leave out something?

The third way was using discriminant

Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$

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

$x^2 -2x+1 = 0$

Discriminant = $(-2)^2 -4(1)(1) = 0$

For $\frac{2x}{x^2 +1} = -1$,

$x^2 +2x+1 = 0$

Discriminant = $(2)^2 -4(1)(1) = 0$

So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.

Is the methods listed correct?Is there any other ways to do it?

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