Let f be a polynomial function defined on non-negative real numbers satisfying f(x)*f(y) + 2 = f(x) + f(y) + f(x*y) for all x, y = 0. Given that f is a one-one function with f(0) = 1 and f'(1) = 2, find the value of f(5).

Question

Let f be a polynomial function defined on non-negative real numbers satisfying f(x)*f(y) + 2 = f(x) + f(y) + f(x*y) for all x, y >= 0. Given that f is a one-one function with f(0) = 1 and f'(1) = 2, find the value of f(5).

Accepted Answer

26. Define g(x) = f(x) - 1. Then g(0) = f(0) - 1 = 0. The functional equation becomes g(x*y) + 1 = (g(x)+1)*(g(y)+1) - g(x) - 1 - g(y) - 1 + 2, which simplifies to g(x*y) = g(x)*g(y). So g is a multiplicative polynomial on [0, inf) with g(0) = 0. The only polynomial solutions are g(x) = xⁿ. Since f'(1) = g'(1) = 2: g'(x) = n*x^(n-1), so g'(1) = n = 2. Hence g(x) = x² and f(x) = x² + 1. Therefore f(5) = 25 + 1 = 26.

Let f be a polynomial function defined on non-negative real numbers satisfying f(x)f(y) + 2 = f(x) + f(y) + f(xy) for all x, y >= 0. Given that f is a one-one function with f(0) = 1 and f'(1) = 2, find the value of f(5).

Solution

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