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

The real cube root in the denominator vanishes at $x=-1$, so the printed rule needs the domain of a function $D_1=\mathbb{R}\setminus \{-1\}$. On its actual domain of a function, the second iterated function simplifies toThis suggestsFor the induction step, assume the formula at $n$ and restrict to arguments for which its value is not $-1$. Its cube is $x^3/(1+n{x}^3)$, so another application givesReal cube roots are multiplicative, including for negative arguments. The newly excluded point satisfies $x^3/(1+n{x}^3)=-1$, hence $x^3=-1/(n+1)$; all earlier exclusions remain. This proves both the formula and the domain of a function by mathematical induction. The inherited exclusions are essential for an iteration of a partial function: cancellation in the final expression cannot repair an undefined intermediate application.

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 $f:\mathbb{R}\setminus \{-1\}\to \mathbb{R}\setminus \{1\}$.