Find its inverse. State the domain and the range of f and f^-1. Graph f, f^-1, and y=x on the same coordinate axes.
f (x)= (x^3) + 1
Unless otherwise stated, the domain of any polynomial is always $\displaystyle \mathbf{R}$.
It always helps to sketch the graph to start with.
So the domain of $\displaystyle f(x) = x^3 + 1$ is $\displaystyle \mathbf{R}$.
It's range is also $\displaystyle \mathbf{R}$.
To find the inverse, your x and y (or in this case, f) values swap, and so do the domain and range.
Since the domain and range of $\displaystyle f$ were both $\displaystyle \mathbf{R}$, so are the domain and range of $\displaystyle f^{-1}$.
Now we just find what the inverse function is.
$\displaystyle x = y^3 + 1$
$\displaystyle x - 1 = y^3$
$\displaystyle y = \sqrt[3]{x - 1}$
$\displaystyle f^{-1}(x) = \sqrt[3]{x - 1}$.
Now graph this function.