Solving x<sin(x)<x w/ Mean Value Theorem

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Homework Help Overview

The discussion revolves around the inequality -x < sin(x) < x and the application of the Mean Value Theorem to demonstrate this relationship.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss attempts to apply the Mean Value Theorem, with one participant expressing confusion over their results involving tan(x). Others question the correctness of the approach and suggest alternative values for a and b in the theorem.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the properties of the cosine function, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are exploring the implications of the maximum and minimum values of the cosine function, as well as the constraints of the problem setup.

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Homework Statement



-x<sin(x)<x

Homework Equations



show the inequality using the mean value theorem.

The Attempt at a Solution


i try to find c but i keep getting tan(x) as the solution.
 
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If you show me what you've tried, I can help you better.
 
i had:

f(x)=sin(x) a=-x b=x

f(x)-f(-x)= f'(c) (x+x)
sin(x) = sin(x)=cos(c) (2x)
2sin(x)=2cos(xc)
tan(x)=c

i don't know if that's right, but i don't get the result.

i would appreciate your help.
 
First, [itex]2x \cos (c) \neq 2 \cos(cx)[/itex] and second [tex]\frac{\cos(cx)}{\cos (x)}\neq c[/tex]!

Try again, but this time use a=0 and b=x.

What do you know about the maximum and minimum values of cosine of any number?
 
ok, now i got:

f(x)=sin(x) a=0 b=x

f(x)-f(0)= f'(c) (x-0)
sin(x) =cos(c) (x)
sin(x)/x=cos(c)

im stucked there...
i don't know what you mean with the the maximum and minimum values of cosine of any number.
 
Well, cosine is a periodic function that is never greater than 1 or less than negative 1...ring a bell?

That means that [itex]-1\leq \cos (c) \leq 1[/itex] and so...
 

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