How do I solve this inverse function?

Question

f(x) = x²-4/x²+4 for x>2 Find the inverse.

This seems like a really simple question but I simply can’t figure it out. Any help would be much appreciated :)

Answer 1

f(x) = (x^2-4)/(x^2+4) = y

to find the inverse switch x and y and solve for the new y

(y^2-4)/(y^2+4) = x

y^2-4 = xy^2 +4x

y^2-xy^2 = 4x +4

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

f^-1(x) = +/-2sqr[(1+x)/(1-x)]

but it's not a function as there are 2 y values for 1 x value

unless one of the y values is an extraneous solution

introduced when multiplying by 1/(1-x)

if not

There is no inverse function

it's an inverse relation

try graphing it and the intercepts are (0,4) and (-1,0) with asymptotes

it's a rectangular hyperbola which fails the vertical line test. It's not function, just an inverse relation

UNLESS you really meant

f(x) = x^2 -4/x + 4

then

x=y^2 -4/y +4

xy = y^3 -4 + 4y

y^3 +(4-x)y -4 = 0

solve for this new y = f^-(x)

but cubic equations are not easily solved and you may introduce an extraneous solution for y

once you multiplied by y to get the cubic equation

try graphing it. It also fails the vertical line test, cutting the graph in 3 places, so even

eliminating one y value, you still don't have a function