# Question on functions and trigonometry.

Question 1

## Homework Statement

Show that, when restricted to the domain [-1,1], g has an inverse function. Use a definition of a theorem, a graph will not be accepted.

## Homework Equations

g(x) = x/(x^2 + 1)

## The Attempt at a Solution

I tried solving this first by going

y = x/(x^2 + 1)
then making x the subject... so..

yx^2 + y = x
yx^2 - x + y = 0

using quadratic formula, i end up with

g^-1(x) = (1+- root(1 - 4x^2) ) / 2x

But I don't get how to prove its an inverse function when x is between -1 and 1.. :(

Question 2...

## Homework Statement

Simplify sin(2tan^-1(x)), tan^-1 is arctan, and state where your simplification is valid.

## The Attempt at a Solution

I made a right angled triangle, with adjacent length 1, opposite length x and hypotenuse length root(1 + x^2).

But I don't know how to deal with that 2...

Thanks (in advance) for helping :D

tiny-tim
Homework Helper
Hi mgnymph!

(have a square-root: √ and a ± and try using the X2 tag just above the Reply box )
Question 1

## Homework Statement

Show that, when restricted to the domain [-1,1], g has an inverse function. Use a definition of a theorem, a graph will not be accepted.

## Homework Equations

g(x) = x/(x^2 + 1)

yx^2 - x + y = 0

using quadratic formula, i end up with

g^-1(x) = (1+- root(1 - 4x^2) ) / 2x

That's inside-out1

You mean x = (1 ± √(1 - 4y2))/2y.

But anyway, wouldn't it be easier just to prove that g(x) is always increasing (by calculus or by a trig substitution)?
Question 2...

## Homework Statement

Simplify sin(2tan^-1(x)), tan^-1 is arctan, and state where your simplification is valid.

Rewrite it …

if y = tanx, then sin2y = … ?

Hi mgnymph!

(have a square-root: √ and a ± and try using the X2 tag just above the Reply box )

That's inside-out1

You mean x = (1 ± √(1 - 4y2))/2y.

But anyway, wouldn't it be easier just to prove that g(x) is always increasing (by calculus or by a trig substitution)?

Rewrite it …

if y = tanx, then sin2y = … ?

oh! Thanks for opening my eyes :D