Using the Mean Value Theorem to Solve for M in f(b)-f(a)/b-a=MHomework Equations

In summary: This is definitely a tough problem. MVT seems like it should work, but it's not clear that it always does. For any given a and b, there might be another c such that f(c) - M<0, but f(b)-M still doesn't equal M.
  • #1
Mathguy15
68
0

Homework Statement


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

Homework Equations



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

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?
 
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  • #2
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
>:( I'm confused now...
 
  • #4
HallsofIvy said:
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.

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.
 
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  • #5
And, for every M>0, there exists x>M such that cos(x)=0...

Graaaaaaaaaaaaaah Calculus!
 
  • #6
Maybe I should read an online book from Mit Online Courseware. These notes I'm reading through are unclear in places >:(
 
  • #7
Mathguy15 said:
Maybe I should read an online book from Mit Online Courseware. These notes I'm reading through are unclear in places >:(

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
algebrat said:
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're having, in what appears to be the difficult job of self motivated study. Are you not taking a class? Where did you get your notes?

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.
 
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  • #9
Mathguy15 said:

Homework Statement


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

Homework Equations



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

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?

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 [itex]\lim_{x \rightarrow 0} f(x) = -\infty[/itex]. 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 [itex]x_0[/itex].

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.
 
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  • #10
Mathguy15 said:
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.

Ugh... this illustrates my point

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

This page makes absolutely no sense... I might go read a bit of the darn book.
 

What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that for any continuous and differentiable function on a closed interval, there exists at least one point in the interval where the slope of the tangent line is equal to the slope of the secant line connecting the endpoints of the interval.

What is the significance of the Mean Value Theorem?

The Mean Value Theorem provides a way to guarantee the existence of a specific point on a curve, which can be useful in solving optimization problems and finding the roots of equations. It also gives insight into the relationship between the graph of a function and its derivative.

How is the Mean Value Theorem used in real-world applications?

The Mean Value Theorem is used in various fields such as economics, physics, and engineering to model and solve real-world problems. For example, it can be used to find the maximum and minimum values of a function, calculate rates of change, and determine the average velocity of an object.

What are the conditions for the Mean Value Theorem to hold?

The Mean Value Theorem holds if the function is continuous on a closed interval and differentiable on the open interval within that closed interval. Additionally, the endpoints of the interval must have the same value, and the function must not have any vertical tangents or corners within the interval.

Can the Mean Value Theorem be generalized to higher dimensions?

Yes, the Mean Value Theorem can be extended to higher dimensions through the use of partial derivatives. In this case, the theorem states that there exists a point in the domain of the function where the directional derivative in the direction of any tangent vector is equal to the average rate of change of the function along that direction.

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