Question

How do you find the inverse of `f(x)= x^3+x` and is it a function?

Answer 1

`g(y) = (y/2 - sqrt[1/27 + y^2/4])^(1/3) + (y/2 + sqrt[1/27 + y^2/4])^(1/3)`

Explanation:

`f(x)` is polynomial, monotonic increasing. This can be stated from

`(df)/(dx)=3x^2+1>0 forall x in RR`

The inverse function `g(x)` is such that

`f(x)=y iff g(y) = x`

The inverse function calculation for

`f(x) = x^3+x =y`

involves the calculation of `x = g(f(x)) = g(y)` and this is obtained using an inversion formula such as Cardanos's
(see https://en.wikipedia.org/wiki/Cubic_function) for resolution of

`x^2+p x +q = 0`. Here `p = 1` and `q = -y`

`x = (-(q/2) + sqrt[(q/2)^2 + (p/3)^3] )^{1/3} +(-(q/2) - sqrt[(q/2)^2 + (p/3)^3])^{1/3}`

giving

`x = (y/2 - sqrt[1/27 + y^2/4])^(1/3) + (y/2 + sqrt[1/27 + y^2/4])^(1/3)`

so
`g(y) = (y/2 - sqrt[1/27 + y^2/4])^(1/3) + (y/2 + sqrt[1/27 + y^2/4])^(1/3)`

The attached figure shows in blue `f(x)` and in red `g(x)`