91Ó°ÊÓ

A function T from X to Y is called invertible if the equation$\mathrm{T}\left(\mathrm{x}\right)=y$ has a unique solution x in X for each y in Y.

In this case, the inverse$T^{-1}$ from Y to X is defined by

$T^{-1}\left(y\right)=(\mathrm{The}\mathrm{unique}x\mathrm{in}X,\mathrm{such}\mathrm{that}T(x)=y)$

To put it differently, the equation

$x=T^{-1}\left(y\right)$means that $y=T\left(x\right)$.

We need to consider the equation$f\left(x\right)=y$for an arbitrary $y∈R$. The given equation$x^3+x=y$is the polynomial equation of odd degree, then it must have at least one solution over the set of real numbers. Further, we need to prove that this solution is unique. Assume that there is a such that the equation$f\left(x\right)=y$have at least two solutions $\mathrm{Î\pm },\mathrm{β}$, then

${\mathrm{Î\pm }}^3+\mathrm{Î\pm }={\mathrm{β}}^3+\mathrm{β}$

That is,

$$\begin{gathered} {\mathrm{Î\pm }}^3-{\mathrm{β}}^3+\mathrm{Î\pm }-\mathrm{β}=0 \\ \left(\mathrm{Î\pm }-\mathrm{β}\right)\left({\mathrm{Î\pm }}^2+\mathrm{Î\pm }\mathrm{β}+{\mathrm{β}}^2\right)+\left(\mathrm{Î\pm }-\mathrm{β}\right)=0 \\ \left(\mathrm{Î\pm }-\mathrm{β}\right)\left({\mathrm{Î\pm }}^2+{\mathrm{β}}^2+\mathrm{Î\pm }\mathrm{β}+1\right)=0 \\ \left({\mathrm{Î\pm }}^2+\mathrm{Î\pm }\mathrm{β}+{\mathrm{β}}^2+1\right)=0 \end{gathered}$$

Our assumption leads to the contradiction that $0=\left({\mathrm{Î\pm }}^2+{\mathrm{β}}^2+a\mathrm{β}+1\right)$but this is not possible because for the arbitrary $\mathrm{Î\pm },\mathrm{β}$expression

${\mathrm{Î\pm }}^2+\mathrm{Î\pm }\mathrm{β}+{\mathrm{β}}^2+1={\left(\mathrm{Î\pm }+\frac{{\mathrm{β}}^2}{2}\right)}^2+\frac{3{\mathrm{β}}^2}{4}+1≻0$

Contradiction!

Therefore, the equation$f\left(x\right)=y$ has unique solution for an arbitrary$y∈R$ then by definition of invertible function in Step 1 we have that the given function is invertible.

Hence, the given function is Invertible.