Question on functions and trigonometry.

AI Thread Summary
The discussion revolves around two mathematical problems involving functions and trigonometry. The first question asks to demonstrate that the function g(x) = x/(x^2 + 1) has an inverse on the domain [-1,1], with suggestions to prove that g(x) is always increasing using calculus or trigonometric substitution. The second question involves simplifying sin(2tan^-1(x)), where participants discuss using a right triangle to derive the simplification. The conversation highlights the importance of understanding inverse functions and trigonometric identities in solving these problems. Overall, the thread emphasizes collaborative problem-solving in mathematics.
mgnymph
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Sorry, for overloading you guys, but I just can't get my head around these!

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
 
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Hi mgnymph! :smile:

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

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)? :wink:
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 = … ? :smile:
 
tiny-tim said:
Hi mgnymph! :smile:

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


That's inside-out1 :rolleyes:

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)? :wink:


Rewrite it …

if y = tanx, then sin2y = … ? :smile:


oh! Thanks for opening my eyes :D
 
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