- #1
mgnymph
- 15
- 0
Sorry, for overloading you guys, but I just can't get my head around these!
Question 1
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.
g(x) = x/(x^2 + 1)
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...
Simplify sin(2tan^-1(x)), tan^-1 is arctan, and state where your simplification is valid.
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
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