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$$f(x)+xf(-x)=x-2$$ what is $f(x$)? I try to solve this problem, but I don't know how to remove $f(-x)$ or converting it to $f(x)$.

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3 Answers 3

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Replacing $x$ by $-x$, we have $$f(-x) - xf(x) = -x - 2$$

Substituting back into the original equation $$f(x) + x(xf(x) -x - 2) = x- 2$$ $$(1+x^2)f(x) = x^2 +3x - 2$$ $$f(x) = \frac{x^2 +3x - 2}{1+x^2}$$

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we have $$f(x)+xf(-x)=x-2$$ setting $x=-x$ we obtain $$f(-x)-xf(x)=-x-2$$ thus we have $$f(x)+x(-x-2+xf(x))=x-2$$ and we get after eliminating $f(x)$ $$f(x)=\frac{x^2+3x-2}{1+x^2}$$

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  • $\begingroup$ "Setting $x=-x$", we obtain $x=0$ and $f(0)+0f(0)=0-2$. Perhaps you meant "replacing $x$ by $-x$" or "setting $y=-x$". $\endgroup$
    – JiK
    Nov 30, 2014 at 11:08
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After solving such task as in the above answers, you should always test it: you put $\frac{x^2 +3x - 2}{1+x^2}$ instead of $f(x)$ into the given equation:

$$\frac{x^2 +3x - 2}{1+x^2} + x\frac{x^2 -3x - 2}{1+x^2} =$$ $$= \frac{x^2 + 3x - 2 +x^3 - 3x^2 - 2x}{1+x^2} =$$ $$= \frac{x^3 - 2x^2 + x - 2}{1+x^2} =$$ $$= \frac{(x-2) (x^2+1)}{1+x^2} = x-2$$

I forgot to do it once on a maths competition and it cost me a point.

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  • $\begingroup$ Why does one needs to test? You should be confident about your calculations. Else you will just loose time in exams. $\endgroup$
    – Babai
    Nov 29, 2014 at 13:55
  • $\begingroup$ @Susobhan,I don't know why,but contests seem to cut points for not checking the validity of solutions of functional equations.Maybe this is because the solution we get may not be a function,I am not sure. $\endgroup$
    – rah4927
    Nov 29, 2014 at 15:27
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    $\begingroup$ Testing is necessary to show that you didn't start with a contradiction: anything logically follows from a contradiction, and the other answers have not shown that the starting point was not a contradiction. $\endgroup$
    – hvd
    Nov 29, 2014 at 20:30
  • $\begingroup$ Or you could see it this way: other answers proved: if <given conditions> then f=... So an implication, not equivalence. So you need to proove the implication the other way, too: if f=... then <given conditions>. The conditions are not necessarily contradictory when it doesn't work. Such tasks are often set as "Find all f that satisfy ...". You might conclude that f belongs to set F, but do all functions from set F satisfy ...? $\endgroup$
    – Heimdall
    Nov 30, 2014 at 23:46

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