### Prove that f(x)=x^3+x has inverse function.

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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).**

You should notice that the function `f(x) = x^3 + x` is continuous over `(-oo,oo)` , hence it is bijective.

You may evaluate its inverse `f^(-1)(x)` , using the following relation between a function and its inverse, such that:

`(f^(-1)(x))' = 1/(f'(x)) => (f^(-1)(x))' = 1/(3x^2 + 1)`

You need to integrate both sides to evaluate `f^(-1)(x)` such that:

`f^(-1)(x) = int 1/(3x^2 + 1) dx`

`f^(-1)(x) = (1/3) int 1/(x^2 + 1 / 3 ) dx`

`f^(-1)(x) = (1/3)*1/(1/sqrt3) tan^(-1) (x/(1/sqrt3)) + c`

`f^(-1)(x) = (sqrt3/3)tan^(-1)(x*sqrt 3) + c`

**Hence, testing if there exists the inverse function, yields that there exists `f^(-1)(x) = (sqrt3/3)tan^(-1)(x*sqrt 3) + c.` **

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