Find the inverse? f(x) =2/x^2+1 (for x ≥ 0)?

May 2021 2 26 Report

Find the inverse? f(x) =2/x^2+1 (for x ≥ 0)?

this is what I have so far:

step 1. y= 2/x^2+1 (for y ≥ 0)

step 2. x= 2/y^2+1 (solve for y)

I m stuck trying to solve for y now. Help is very much appreciated.

y = 2 / (x^2 + 1), for x ≥ 0 the range is 0 < y ≤ 2

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

x * (y^2 + 1) = 2

y^2 x + x = 2

y^2 x = 2 - x

y^2 = (2 - x)/x

y = ± √[(2 - x)/x]

y = √[(2 - x)/x] , domain: 0 < x ≤ 2, range: y ≥ 0

Is that f(x) =2/(x^2+1) or f(x) =(2/x^2) + 1?

Assume PEMDAS, so the second option.

x = 2/y^2 + 1

x-1 = 2/y^2

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

y^2 = 2/(x-1)

y = ±√(2/(x-1)