f(x) = x^3+x.

f(x) = x^3+x is defined for all real values of x whose domain is the set of all real values that x takes.The range of the function is also the set of all real values. So f(x) is an onto function.

Also f(x) = x^3+x is an one-one function as different elements of the domain has different images. That is, if x1 and x2 are different, then f(x1) and f(x2) are also different.

Suppose for x = a and x = b and a is not = b.

Then let f(a) = f(b).

f(a) = a^3+a by definition of the function.

f(b) = b^3+b by definition of the function.

So by assumption f(a) = f(b) => a^3+a= b^3+b.

=> a^3-b^3 + (b-a) = 0.

=> (a-b){ a^2+ab+b^2+1} = 0

=> a = b.

Thus f(x) = x^3+x is an onto and one-one function. Therefore f(x) = x^3+x has an inverse function.

For the given function to have an inverse function f^-1(x), it has to be bijective.

For the function to be bijective, it has to be one to one function and on-to function.

We'll prove that the given function is one to one function.

For this reason, we'll have to prove that the function is strictly increasing or decreasing. For this reason, we'll determine the first derivative of the function.

f'(x) = (x^3+x)'

f'(x) = 3x^2 + 1

It is obvious that f'(x)>0 for any value of x, so the function f(x) is an one to one function.

Let's prove that the function is an on to function. We'll determine the limits of the function for x-> +infinite and x->-infinite

Lim f(x) = lim (x^3 + x) = (+infinite)^3 + infinite = +infinite

Lim f(x) = lim (x^3 + x) = (-infinite)^3 - infinite = -infinite

So, f(x) is a continuous function, then it is an on-to function.

**Since the fuction is both, one to one and on to function, it is bijective and it does exist f^-1(x).**