How to find the inverse of a rational function when the polynomials are higher than degree 1?

Question

f(x) = (3x) / (x² + 1). Show that f^-1(x)  = (6 - 3x²) / (x² + 2)²

Answer 1

y = (3x)/(x^2 + 1)


Switches X and Y


x = (3y)/(y^2 + 1)

Solves for y:

(y^2 +1)x - 3y = 0

x*y^2 - 3y + x = 0

Quadratic formula with A=x , b= -3, C = x

y = [3 +or- sqrt( 9 - 4x^2) ]/(2x) = f-inverse(x)

To prove that this is the correct answer, it must be
shown that f ( f-inverse (x)) = x and

f-inverse ( f ( x) ) = x

That is when f and f-inverse are composed, the result
is the identity line x. It must work in both directions,
f comp f-inverse and f-inverse comp f


Plugging f-inverse into f:
-----------------------------

[ f-inverse ]^2 = (9 +or- 6*sqrt(9-4x^2) + 9 - 4x^2 )/(4x^2)

= (18 - 4x^2 +or- 6 * sqrt(9-4^2))/(4x^2)

= (9 - 2x^2 +or- 3 * sqrt(9-4x^2))/(2x^2)

Adding 1 to this causes the x^2 term to drop out, as the common denominator is 2x^2.
It becomes:

[9 +or- 3 * sqrt(9-4x^2)]/(2x^2)

The reciprocal is (2x^2)/[9 +or- 3 * sqrt(9-4x^2)] <--- we have found 1/(z^2+1) where z=f-inverse;
call this expression ALPHA

3 * f-inverse = 3[ 3 +or- sqrt( 9 - 4x^2) ]/(2x)

= (3/2) [ 3 +or- sqrt( 9 - 4x^2) ]/x

Next Multiplies this by expression for 3 * f-inverse by ALPHA, cancelling the
bracketed expression containing the square root by a factor of 3

(2x^2) * (3/6x) = 6x^2/6x = x


-------------------------------------------------------------------------------------------

Plugging f into f-inverse:

Squaring the original function, [f]^2 =(9x^2)/(x^2+1)^2

Multiplying that by -4 : -36x^2/(x^2+1)^2

Adding 9, [9(x^2+1)^2 - 36x^2 ]/(x^2+1)^2

[ 9(x^4 + 2x^2 +1) - 36x^2]/(x^2 +1)^2

[ 9x^4 + 18x^2 + 9 - 36x^2 ]/(x^2+1)^2

[ 9x^4 - 18x^2 + 9]/(x^2+1)^2 =

9 (x^4 - 2x^2 + 1)/(x^2 +1)^2 =
9 (x^2 -1)^2 / (x^2 + 1)^2 =

9 (x+1)^2(x-1)^2/(x^2+1)^2

THe square root of that is 3(x+1)(x-1)/(x^2+1)


So the numerator becomes: 3 +or- 3(x+1)(x-1)/(x^2+1)

3(x^2+1) +or- 3(x+1)(x-1)
--------------------------
x^2 +1

3x^2 + 3 +or- 3x^2 - 3
------------------------
x^2 +1

3x^2 +or- 3x^2
-----------------
x^2 + 1

which is either: 6x^2/(x^2+1) or zero

The latter case forces all functions to vanish and the identity holds by default.


Twice the original function is (6x)/(x^2+1)

Dividing expression BETA by this causes x^2+1 to cancel.
6x^2/6x = x.

Inverse proven.