Solving Injective Function
The injective function $ f(x) $ satisfies
\[f(x) f(x + y) = f(2x + y) - xf(x + y) + x\]for all real numbers $ x $ and $ y $. Find $ f(x) $.
Note: A function $ f $ is injective if $ f(a) = f(b) $ implies $ a = b $.
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- $\frac{a}{b}$
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- $\infty$
Solution
Setting $ x = y = 0 $ in the given functional equation, we get
\[f(0)^2 = f(0),\]so $ f(0) = 0 $ or $ f(0) = 1 $.
Setting $ x = 0, $ we get
\[f(0) f(y) = f(y).\]If $ f(0) = 0, $ then $ f(y) = 0 $ for all $ y, $ but this function is not injective. Hence, $ f(0) = 1 $.
Setting $ y = x, $ we get
\[f(x) f(2x) = f(3x) - xf(2x) + x\]for all $ x $.
Setting $ x = 2t $ and $ y = -t, $ we get
\[f(2t) f(t) = f(3t) - 2tf(t) + 2t\]for all $ t $. In other words,
\[f(2x) f(x) = f(3x) - 2xf(x) + 2x\]for all $ x $. comparing this to the equation $ f(x) f(2x) = f(3x) - xf(2x) + x, $ we can conlucde that
\[-xf(2x) + x = -2xf(x) + 2x,\]or $ xf(2x) = 2xf(x) - x $ for all $ x $. Assuming $ x $ is nonzero, we can divide both sides by $ x, $ to get $ f(2x) = 2f(x) - 1 $. Since this equation holds for $ x = 0, $ we can say that it holds for all $ x $.
Setting $ y = 0, $ we get
\[f(x)^2 = f(2x) - xf(x) + x\]Substituting $ f(2x) = 2f(x) - 1, $ we get
\[f(x)^2 = 2f(x) - 1 - xf(x) + x,\]so
\[f(x)^2 + (x - 2) f(x) - x + 1 = 0.\]This factors as
\[(f(x) - 1)(f(x) + x - 1) = 0.\]Hence, $ f(x) = 1 $ or $ f(x) = 1 - x $ for each individual value of $ x $. If $ x \neq 0, $ then $ f(x) $ cannot be equal to 1, since $ f $ is injective, so $ f(x) = \boxed{1 - x} $. Note that this formula also holds when $ x = 0 $.
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