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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.

pie
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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
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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. – Mark Bennet Dec 29 '12 at 20:40

1 Answers1

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

Nameless
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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
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    it is either 0 or negative, while $g(5)$ is either 0 or positive. – pie Dec 29 '12 at 20:42
  • @pie Exactly. Is $g$ continuous? – Nameless Dec 29 '12 at 20:42
  • 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. – pie Dec 29 '12 at 20:47
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    @pie. No. The continuity of $g$ follows from the continuity of $f$ and $2x-3$ – Nameless Dec 29 '12 at 20:49
  • 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 :/ – pie Dec 29 '12 at 20:56
  • @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? – Nameless Dec 29 '12 at 20:57
  • Ah.. I got it now. Sorry for being so dumb, and thanks! – pie Dec 29 '12 at 20:58
  • @pie Sure. My advice: Just concentrate on what the problem is, what you want to show and never lose sight of that. – Nameless Dec 29 '12 at 21:00