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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?2012-12-29
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    @MarkBennet By one I suppose he means at least one, as more roots could exist.2012-12-29
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    @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.2012-12-29

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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?2012-12-29
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    Sorry.. I think I'll have to be spoon-fed here. I don't think I graspt the theory in this one.2012-12-29
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    @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 $g(-2)=f(-2)-7$?2012-12-29
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    it is either 0 or negative, while $g(5)$ is either 0 or positive.2012-12-29
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    @pie Exactly. Is $g$ continuous?2012-12-29
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    So we get that $g(5)\ge0$ and $g(-2)\le0$. That means it is continous because they both might be equal 0? Sorry, I'll be happy if you cleared that one out.2012-12-29
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    @pie. No. The continuity of $g$ follows from the continuity of $f$ and $2x-3$2012-12-29
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    Ok, so it has to do now with the intermidiate value theorom that there is a value between $g(5)$ and $g(-2)$ which equals $f(5)$ or $f(-2)$? That's the direction...? If not then I'm kinda lost :/2012-12-29
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    @pie Ok let me go over this. What do we want to show? That your equation has a solution right? This means that $g(x_0)=0$ for some $x_0\in \mathbb{R}$. Do you follow?2012-12-29
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    Ah.. I got it now. Sorry for being so dumb, and thanks!2012-12-29
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    @pie Sure. My advice: Just concentrate on what the problem is, what you want to show and never lose sight of that.2012-12-29