 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/1/e/i/solution

Replace

x

1/ x

, which stays in the domain of the functional equation. Writing

u = f (x)

and

v = f (1/ x)

u + 2 v = x, 2 u + v = \frac{1}{x} .

Twice the second equation minus the first gives

3 u = 2/ x - x

, so the only possible function is

f (x) = \frac{2/ x - x}{3} (x > 0) .

To check existence as well as uniqueness, substitute this expression:

f (x) + 2 f (1/ x) = (2/ x - x + 4 x - 2/ x) /3 = x

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