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


by nuadre
Tags: prove, rolle, theorem
nuadre
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#1
Jun8-10, 01:03 AM
P: 6
1. The problem statement, all variables and given/known data

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


2. Relevant equations

-

3. 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)

------------------------------

Do i substitute values in for a,b,c to try to prove? I'm so horribly confused :(
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Tedjn
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#2
Jun8-10, 01:07 AM
P: 740
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.
nuadre
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#3
Jun8-10, 01:52 AM
P: 6
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.

Tedjn
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#4
Jun8-10, 02:17 AM
P: 740

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


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?
nuadre
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#5
Jun8-10, 03:18 AM
P: 6
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."
-------------------------------------
"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?
Tedjn
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#6
Jun8-10, 06:45 AM
P: 740
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.
nuadre
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#7
Jun8-10, 11:38 PM
P: 6
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?
Tedjn
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#8
Jun9-10, 06:44 AM
P: 740
Rolle's theorem doesn't only apply to straight, horizontal lines.
LCKurtz
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#9
Jun9-10, 11:13 AM
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Quote Quote by nuadre View Post
1. The problem statement, all variables and given/known data

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

Try applying Rolle's theorem to

[tex]F(x) = \left| \begin{array}{ccc}
1 & a & f(a)\\
1 & b & f(b)\\
1 & x & f(x)
\end{array}\right|[/tex]


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