Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/5/b/ii/solution

For , solve with . Multiplying by the positive denominator gives , so the quadratic formula gives
The minus sign gives a negative number and is inadmissible. The plus sign gives a positive number for every , and it is the unique positive solution. Hence the inverse function is
Indeed , so . Conversely, for , , giving . These verify both inverse function identities.

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