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Exercise 16Calculus › Single-variable calculus › Continuity › Functional equationsL1Added 4 Nov 2026

f(x)f(y) − f(xy) = x + y

#substitution#special values#verification

Statement

Find all functions f:R→Rf:\mathbb{R}\to\mathbb{R} such that for all real x,yx,y,
f(x)f(y)−f(xy)=x+y.f(x)f(y)-f(xy)=x+y.

Hint

Try y=0y=0: the term f(xy)f(xy) becomes a constant.

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Solution

Substitution y=0y=0. f(x)f(0)−f(0)=xf(x)f(0)-f(0)=x, i.e.
f(0)(f(x)−1)=xfor all x.f(0)\bigl(f(x)-1\bigr)=x\quad\text{for all }x.
If f(0)=0f(0)=0 the left side is 00 for all xx, which is absurd (take x=1x=1). So f(0)≠0f(0)\ne0 and
f(x)=1+xf(0).f(x)=1+\frac x{f(0)}.

Determining the constant. Setting x=0x=0 in this formula gives f(0)=1f(0)=1. Hence f(x)=x+1f(x)=x+1.

Answer. f(x)=x+1f(x)=x+1 is the only solution.

Lesson. Always: (1) try x=0x=0, y=0y=0, x=yx=y, y=1y=1; (2) when an equation is solved by a formula, plug the formula back to check it is a solution.