Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 1 / f / ii

Set $y=f(x)$ and cube the equation. Solving for $x^3$ givesConsequently the inverse function isThe value $y=1$ is impossible in the original equation. Conversely, every real $y\ne 1$ gives exactly one real $x$ by the displayed cube root, and this $x$ cannot equal $-1$. Direct substitution verifies both inverse identities. Thus the corrected bijection is $f:\mathbb{R}\setminus \{-1\}\to \mathbb{R}\setminus \{1\}$.