 Solution

ID: 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/solution

y = f (x)

and cube the equation. Solving for

x^{3}

gives

y^{3} = \frac{x ^{3}}{1 + x ^{3}}, x^{3} = \frac{y ^{3}}{1 - y ^{3}} .

Consequently the inverse function is

f^{- 1} (y) = \frac{y}{3 1 - y ^{3}}, y \neq = 1.

The value

y = 1

is impossible in the original equation. Conversely, every real

y \neq = 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 : R ∖ {- 1} \to R ∖ {1}

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