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Question 15

The real valued function $$f(x)$$ satisfies the equation $$2f(1-x)+1=xf(x)$$ for all $$x$$. Then $$\left(x^2-x+4\right)f(x)$$ equals

Solution

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Replace $$x$$ by $$1-x$$ to obtain $$2f(x)+1=(1-x)f(1-x)$$. From the original equation, $$f(1-x)=\frac{xf(x)-1}{2}$$. Substitution gives $$4f(x)+2=x(1-x)f(x)-(1-x)$$, so $$\left(x^2-x+4\right)f(x)=x-3$$.

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