# Mean Value Theorem Question

1. ### Mathguy15

63
1. The problem statement, all variables and given/known data
Let f(x)=log(x)+sin(x) on the positive real line. Use the mean value theorem to assure that for all M>0, there exists positive numbers a and b such that f(b)-f(a)/b-a=M

2. Relevant equations

f'(x)=1/x+cos(x)

3. The attempt at a solution

I know that as x→0, f'(x) gets arbitrarily large and as x→∞, f'(x) gets arbitrarily small, so for every M>0 there exists a and b such that f'(a)-M<0 and f'(b)-M>0. f'(x)-M would then have a root by the intermediate value theorem. From here, I don't know where to go. I know that as h→0, f(r+h)-f(r)/h→f'(r), where r is the root of f'(x)-M, so as h→0, the intervals (r,r+h) are such that f(b)-f(a)/b-a gets arbitrarily close to M by the Mean Value Theorem, but I don't know where to go. Any ideas?

2. ### HallsofIvy

40,785
Staff Emeritus
You appear to know what is not true! f'(x) does NOT get "arbitrarily small". For any M> 0, the exist x> M such that cos(x)= 1. In that case, f'(x) will be close to 1, not 0.

3. ### Mathguy15

63
>:( I'm confused now...

4. ### algebrat

428
By the same argument, it get's close to -1...

It is not "not true".

But you are right, we should be this careful. It is like the difference between an unbounded function and one that converges to infinity.

It does attain arbitrarily small values, but it does not converge to zero. To say f'(x) got arbitrarily small did have some ambiguity, which you are right to tear apart like Freddy Krueger.

Last edited: Jul 1, 2012
5. ### Mathguy15

63
And, for every M>0, there exists x>M such that cos(x)=0....

Graaaaaaaaaaaaaah Calculus!

6. ### Mathguy15

63
Maybe I should read an online book from Mit Online Courseware. These notes I'm reading through are unclear in places >:(

7. ### algebrat

428
I see no reason that your notes are necessarily bad, you could take it as HallsOfIvy was giving you/your notes a hard time, but don't necessarily give up on them. Feel free to try more resources, but don't let it unnecessarily stop any momentum you have, in what appears to be the difficult job of self motivated study. Are you not taking a class? Where did you get your notes?

8. ### Mathguy15

63
http://www.math.harvard.edu/~knill/teaching/math1a_2011/handouts.html

Well uh.. this wasn't necessarily due to HallsofIvy. The notes have definitions that differ from other resources (as HallsofIvy has pointed out before), and the answers to homework problems aren't always there (which is why I'm here). The answers to examples also cause one to read between the lines, leaving me with only the foggiest notion of what a limit actually is(though I think I've come up with a definition, but its very long). I think something like an online book(like the one I've found) could help me more, but I won't have much time over the summer....

EDIT:And the notes correspond to a book, which is probably why it is unclear in places.... the book can handle what the student's don't understand.

EDIT:Well, perhaps foggiest isn't the right term. For example, I can calculate using the chain rule and product rule, and I understand their proofs with the very basic intuitive definition of the limit and derivative, but I still feel like I'm missing out bigtime. Given the fact that summer's almost over, I might just tough it out and check in with the book every now and again if I get stuck, assuming the book has a section related to what I'm learning at the time.

Last edited: Jul 1, 2012
9. ### Curious3141

2,970
I don't know if MVT is necessarily the best way to prove this. In fact, I think you need a converse MVT to prove it, and I don't think the converse of the MVT holds in general.

This is far more simply proved using IVT (Intermediate Value Theorem). First observe that $\lim_{x \rightarrow 0} f(x) = -\infty$. Also observe that there is a local maximum value for f(x) around x = 2. Proving this is easy by observing the behaviour of f'(x) in the neighborhood of x = 2. Calculate f'(2) and f'(2.1) and apply IVT to f'(x) to prove that there has to be a stationary point between these two x-values, which is immediately seen to be a local maximum. Call the x-value at that local maximum $x_0$.

Now, consider secant lines originating from x in the open interval (0,x0). Clearly, the secant lines have gradients in the open interval (0,∞). Since the curve y = f(x) is continous, the IVT guarantees that you can always choose a positive x value less than x0 to give you any positive gradient for the secant terminating at x0. Hence, simply letting b = x0 and letting a be a particular positive x smaller than x0 will suffice to complete the proof.

Last edited: Jul 1, 2012
10. ### Mathguy15

63
Ugh... this illustrates my point

http://www.math.harvard.edu/~knill/teaching/math1a_2011/handouts/27-catastrophe.pdf