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Let $f(x)$ be a continues function for all $x$, and $|f(x)|\le7$ for all $x$.

Prove the equation $2x+f(x)=3$ has one solution.

I think the intermediate value theorem is key in this, but I'm not sure of the proper usage.

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The intermediate value theorem might tell you that a solution would exist, but it wouldn't tell you how many - are you looking for at least one solution, or exactly one solution? – Mark Bennet Dec 29 '12 at 20:31
@MarkBennet By one I suppose he means at least one, as more roots could exist. – Nameless Dec 29 '12 at 20:34
@Nameless I assume so too, but I was interested in whether the question had been correctly posed (there might be a monotonicity criterion missed) and in pointing out an ambiguity in the way the question is asked, which might help the person who asked the question to formulate their thoughts more clearly. – Mark Bennet Dec 29 '12 at 20:40
up vote 3 down vote accepted

Hint: Let $g(x)=f(x)+2x-3$. Then $g(-2)$ and $g(5)$ are...

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So if g(-2)+g(5)=f(-2)+f(5), g(x)=f(x)? Just so I'll get it.. If I find an x value (in this case 1.5) independent of f(x) then it proves the equation has at least one solution? – pie Dec 29 '12 at 20:36
Sorry.. I think I'll have to be spoon-fed here. I don't think I graspt the theory in this one. – pie Dec 29 '12 at 20:39
@pie Ok. You have $\left|f(x)\right|\le 7\iff -7\le f(x)\le 7$. What does this tell you about the sign of $g(-2)=f(-2)-7$? – Nameless Dec 29 '12 at 20:40
it is either 0 or negative, while $g(5)$ is either 0 or positive. – pie Dec 29 '12 at 20:42
@pie. No. The continuity of $g$ follows from the continuity of $f$ and $2x-3$ – Nameless Dec 29 '12 at 20:49

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