How do you use Rolle's Theorem to Prove the Mean Value Theorem?

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Rolle's Theorem states that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists at least one point c in (a,b) where f'(c) = 0. The Mean Value Theorem (MVT) generalizes this by stating that if a function is continuous on [a,b] and differentiable on (a,b), there exists a point c such that (f(b) - f(a)) / (b - a) = f'(c). To prove the MVT using Rolle's Theorem, one can consider a modified function g(x) = f(x) - h(x) that meets the conditions of Rolle's Theorem. The discussion emphasizes that while specific examples can illustrate the concepts, a general proof is necessary to establish the validity of the MVT. Understanding that Rolle's Theorem is a special case of the MVT is crucial for grasping the relationship between the two theorems.
nuadre
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Homework Statement



Assuming Rolle's Theorem, Prove the Mean Value Theorem.


Homework Equations



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The Attempt at a Solution



I know these definitions:

Rolle's Theorem:
If y=f(x) is continuous on all points [a,b] and differentiable on all interior points (a,b),
and if f(a) = f(b)
then there is at least one point c such that f '(c)=0

Mean Value Theorem:
y:f(x) continuous on [a,b], differentiable on all (a,b) then one point such that

f(b) - f(a) / b-a = f ' (c)

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Do i substitute values in for a,b,c to try to prove? I'm so horribly confused :(
 
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Of course in the MVT it's not necessarily true that f(a) = f(b). Can you subtract a natural function from f(x), let's call this h(x), so that g(x) = f(x) - h(x) satisfies g(a) = g(b)? Then, you will be able to apply Rolle's theorem.
 
hi Ted! Thanks for your reply, I couldn't get my head around this.

Can I use values of a= -1 , b= 1, c=0 and sub these in for a function
y= x^2 ?
y' = 2x
so

y(-1) = f(a)
y(1) = f(b)
y'(0) =0
therefore f(b) - f(a) / b-a = f ' (c) is equal to
1-1 / 1+1 =0 =f'(c) ?

Am I on the right track? How would I word this for an answer though ha.
 
No, the MVT is true for all intervals [a,b] and all functions f(x). Substituting values for a and b and a special form f(x) = x^2 is only a special case of MVT. It doesn't prove it in general. First, do you understand why Rolle's theorem is a special case of MVT?
 
Proof of Rolle's Theorem:
"From the extreme value theorem, the function attains its extreme values on [a,b]. If it attains them both at a and b, then the function is constant, and so has zero derivative everywhere. If it attains either of them at an interior point, then by the extreme value derivative theorem the derivative at that point is zero."
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"From the extreme value theorem, the function attains its extreme values on [a,b]. If it attains them both at a and b"

Is this saying it obtains both its global max/global min at both points a and b? So I would be visualising a straight line parallel to the x axis?

"If it attains either of them at an interior point, then by the extreme value derivative theorem the derivative at that point is zero"

I'm now visualising a point c, situated between a and b. If there is a global max/min here then that means the derivative of c = 0?
 
Yes, but explain how it is a special case of MVT. That is, given MVT, why is it almost immediately obvious that Rolle's theorem is true.
 
it's a special case of MVT because at every point between a,b inclusive, the derivative is 0 (As it is a constant straight line parallel to x axis)? Which proves that rolle's theorem is true?
 
Rolle's theorem doesn't only apply to straight, horizontal lines.
 
nuadre said:

Homework Statement



Assuming Rolle's Theorem, Prove the Mean Value Theorem.

Try applying Rolle's theorem to

F(x) = \left| \begin{array}{ccc}<br /> 1 &amp; a &amp; f(a)\\<br /> 1 &amp; b &amp; f(b)\\<br /> 1 &amp; x &amp; f(x)<br /> \end{array}\right|
 

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