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Again, I apologize for what looks like a very narrow question. But there's possibly a general principal at work here that I'm not grasping.

I understand the answer provided for exercise 3 in chapter 7 of Spivak's Calculus (4E, p.130), but wonder if another approach might work (and be closer to the spirit of the chapter). For example for (ii), to show that $$\sin x = x-1$$ has a solution, can't I take $$f(x)=\sin x - x+1$$ and argue that for large $|x|$, $x>0 \Rightarrow f(x)<0$ and $x<0 \Rightarrow f(x)>0$, so that I can use Theorem 1 (p. 122: $f(x)$ continuous on $[a,b]$ and $f(a)<0 )?

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    You certainly *can* apply IVT and it is commonly done. But in your case, why is it evident that $f(x) > 0$ for $x < 0$? It's true, but not immediately obvious. Moreover you don't have to worry about all $x < 0$ or $x \leq 0$ for the IVT argument to go through. It suffices to find one $x$ such that $f(x) > 0$ and $x=0$ works!2011-09-13
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    I was going for generality, partly because of the preceding part of the same problem where $$f(x)=x^{179}+\frac{163}{1+x^{2}+\sin^{2}x}-119,$$which seemed easier to solve using a similar approach.2011-09-13
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    @Sri, $-x\le f(x)\le2-x$ hence $f(x)<0$ for $x>2$ and $f(x)>0$ for $x<0$. Hence "*for large $|x|$*" might mean "*for $x>2$ or $x<0$*" here.2011-09-13
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    @Didier Oh, I see. :) I took that the large $|x|$ applies only for positive $x$. Yes, if the OP meant what you are saying (and I am now sure s/he did), then s/he is absolutely correct.2011-09-13

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I am currently teaching a course from Spivak's Calculus, and I think your solution to this problem is entirely correct.

You considered the auxiliary function $f(x) = \sin x - x + 1$ and showed that it is continuous and takes negative values for sufficiently large, positive $x$ and also positive values for sufficiently large, negative $x$. So by IVT it must take on the value $0$.

You can be a little more explicit about the sufficiently large business though. For instance, you know that $-1 \leq \sin x \leq 1$ for all $x$, so

$-x \leq f(x) \leq -x + 2$.

From this one sees that $f(x)$ is non-negative for all $x \leq 0$ and non-positive for all $x \geq 2$. In this sort of problem, the more complicated the function gets, the more you want to call on "general principles" in order to give you the estimates you need. For instance, if you had a very complicated polynomial $p(x)$ of degree $19$ in place of $x$, you probably don't want to give explicit values but just use the fact that as $x$ approaches $\pm \infty$, so does $p(x)$, while $\sin x$ stays bounded.

Final comment: to be sure, you don't have to find explicit values for this problem. But in order to be best understood you should probably say something in the way of justification of what happens for sufficiently large $x$. Note that in the above paragraph I gave a less explicit answer for a more general class of problems, and as you know there are other problems in the text which are like the one I made up above. But -- and this is an issue of effective mathematical writing and communication rather than mathematical correctness -- there is a sort of principle of economy at work here. In order to be best understood, it's generally a good idea to use the simplest arguments you can think of that justify a given claim, and a lot of people find more explicit arguments to be simpler than less explicit ones. Anyway, not here but sometimes you do have to be explicit and concrete, so it's a good idea to cultivate the ability to do so...

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    Yes, precisely (on all points)! I was perhaps overgeneralizing given the simplicity of the problem at hand, but (again, in the spirit of the chapter) I was trying to come up with an approach that would work more generally (having been intimidated by the first part of the problem). Thanks!2011-09-13
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    @Pete: While I'm at it: is it SO-kosher to ask this sort of "did I understand the problem" or "is my answer right too" question about specific textbook problems? It strikes me that many problems (esp. those in excellent texts like Spivak's) are great jumping off points for important discussions and illustrate significant themes and concepts of the sort that have a place here. Rgds.2011-09-13
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    @rax: you wrote "SO", but I'm assume your talking about math.se. (Anyway, I'm not qualified to answer cultural questions about SO...) Sure, your question is A-OK for this site. Most homework questions are acceptable here, provided the OP shows some level of due diligence at trying to solve them herself. Posting and asking about a correct solution is maximally diligent! And I also agree that Spivak's *Calculus* provides many opportunities for good discussions on this site. In fact, it turns out that my top-voted question here is closely related to a double-starred problem in that text.2011-09-13
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Your approach is correct, but the proof is not water-tight. By which I mean that to complete the proof, one needs the following additional proposition:

Suppose a continuous function $f : \mathbb R \to \mathbb R$ satisfies $f(x) \to \infty$ as $x \to \infty$, and $f(x) \to -\infty$ as $x \to -\infty$. Then there exists some $x_0 \in \mathbb R$ such that $f(x_0) = 0$.

From your question and comments, I feel that you intuitively understand the above statement, and in fact, you are also implicitly using it. However, I am not sure that you know how to prove it. The proof is just a single line, by the way. Nevertheless, I encourage you to do it.

I am emphasizing this point because when one is learning a subject, it is of great pedagogical value to sit down and prove even seemingly obvious statements. Moreover, while one certainly cannot write completely rigorous proofs all the time, one should at least be aware of the little missing details, and also how those details can be taken care of.

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    That's a good point. But that proposition is not (I think) needed for the the argument I was making, namely that as $|x| \to \infty$, $f(x) \to -x$; so for such $x<0, f(x)>0$ and for such $x>0, f(x)<0$. I could easily be missing something though.2011-09-14
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    @rax I don't see what your argument is then. Ok, I get it. You do not need that $f(x)$ grows unbounded. You just need to show that there is some $x$ such that $f(x)>0$ and another $x$ such that it is negative. But how are you showing this?2011-09-14
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    @ Srivatsan: I may not be, so bear with me. But haven't I done so by showing that $f(x) \to -x$ for some $x<0$ and for some $x>0$?2011-09-14
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    @rax Ok, I was waiting for you in the chat room. Anyway.. Your statement that $f(x) \to -x$ doesn't mean anything. How do you define that formally?2011-09-14
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    @ Srivatsan: That depends on what's given. In the context of the exercise, I'm not sure what more I'd need to provide. Perhaps that $$\mathop{\lim}\limits_{|x| \to \infty}\frac{f(x)}{-x} =1.$$ But I may be missing your point.2011-09-14
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    @rax Well, that's at least a meaningful statement. How do you go from here to applying IVT? Even if you are not using *exactly* what statement I had written there, you are using something like it. My point is: it may be worthwhile to identify the precise argument and formalize the proof, **at least once**, before you are fully convinced. (Although at this point, I will say this: your proof looks fine, and will be interpretted correctly by any expert. So I am just nitpicking, if you want to see it that way.)2011-09-14
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    Nitpicking is fine. I _believe_ I can go directly from this to $$\mathop{\lim}\limits_{\lvert x \rvert \to \infty}f(x) =-x,$$ and from there to the steps in my original solution.2011-09-15
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    Two issues with your final comment. (1.) Your last statement is *meaningless* (at least as far as standard definitions go). (2.) I do not see any way of going from this statement to IVT without using some proposition like what I said.2011-09-15
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    How about $$\mathop{\lim}\limits_{x \to \infty}x(\frac{1}{x}\sin x-1+\frac{1}{x})$$ which, since $\sin x$ is in $[-1,1]$ so that $(1/x)\sin x$ goes to $0$ is $$\mathop{\lim}\limits_{x \to \infty}x(0-1+0)=-x$$ so that $$\mathop{\lim}\limits_{x \to \infty^+}f(x)=-\infty; \mathop{\lim}\limits_{x \to \infty^-}f(x)=\infty$$2011-09-19
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    Your steps are correct. Still that leaves (2.) unresolved. Suppose I told you that $f(x)$ goes to $\pm \infty$ as $x \to \mp \infty$. Does this mean that $f$ has a root somewhere? Yes. Why? IVT. But the IVT cannot be applied directly; you should massage it a little bit, and prove some convenient form of it like I did in my answer. Then you can correctly conclude that $f$ has a root.2011-09-19
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    Ok, my bad. I was sloppy in checking. :-/ The final conclusion is correct, but there is a definite trouble in the statement before the final one. @rax2011-09-19
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    Ok, this is the problem. A limit of a function is a number (or sometimes we even tolerate infinity, as a way of saying that the function diverges to infinity). It cannot be another function, like $x$. (I pointed to this several comments before:) Statements like $\lim_{x\to \infty} f(x) = g(x)$ or $f(x) \to g(x)$ as $x \to \infty$ are all inaccurate.2011-09-19
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    Yes, point well taken; though I thought I spotted that [in my other post](http://math.stackexchange.com/questions/65908/is-there-an-expression-for-a-number-always-between-that-is-used-within-expressi) and fixed it, but apparently not.2011-09-19
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    @rax Apologies for my lapse. :)2011-09-19
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    No problem. I was just confused because I thought I hadn't, in fact, corrected it [over there](http://math.stackexchange.com/q/65908/12400). Rgds.2011-09-19
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Perhaps I'm missing something (given the comments posted thus far), but I don't see what is wrong with your argument, other than being a little more precise in applying Theorem 1, such as:

$$f(3) = \sin(3) - 3 + 1 \leq 1 - 3 + 1 = -1 < 0$$

and

$$f(-1) = -\sin(1) - (-1) + 1 \geq -1 + 1 + 1 = 1 > 0,$$

so there exists $x \in [-1,3]$ such that $f(x) = 0$.

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    My argument was the other way round: instead if picking specific values, can't I just say that it's so for large $|x|$?2011-09-13
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    @raxa, if you just say "for large enough $x$", you're in effect _either_ promising to exhibit a concrete bound such that $x$ larger than that bound will always work (which is a bit more cinvolved than just showing a single $x$ that works), _or_ you're leaving it to the reader to convince himself that there is such a bound. The latter is quite defensible in this case, but in principle what you get out of it here is not a simpler proof, but merely a more compact _presentation_ of a slightly less simple proof.2011-09-13
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    @Henning: Yes, and just to be sure I get that point, Pete's demonstration that $f(x)$ is non-negative for all $x \leq 0$ and non-positive for all $x \geq 2$ is an example of how to show such concrete bounds (within the context of an approach that's still "general").2011-09-13