Let y = 1/((3-x)(x+1))

Substitute y with x and x with y.

1 ≤ y < 3

x = 1/((3-y)(y+1))

1/x = (3 - y)(y + 1)

1/x = 2y - y² + 3

y² - 2y - 3 = -1/x

y² - 2y + 1 = -1/x + 4

(y - 1)² = 4 - 1/x

y - 1 = √(4 - 1/x)

( For the inverse, its range corresponds to the domain of g. So 1 ≤ y < 3

Since 1 ≤ y < 3 , this means 0 ≤ y - 1 < 2 . So √(4 - 1/x) ≥ 0 and that's why we don't have -√(4 - 1/x) )

y = √(4 - 1/x) + 1


Therefore,

g‾¹(x) = √(4 - 1/x) + 1



When x = 1, g(x) = 1/((3 - 1)(1 + 1)) = ¼
When x increases from 1 and tends to 3, g(x) increases from ¼ and tends to infinity

So the range of g is ［¼,∞)

The range of g corresponds to the domain of g‾¹ . So,


g‾¹ : x ↦ √(4 - 1/x) + 1, x ≥ ¼