Mean value theorem and indefinate intgral

[wombat4000]

Homework Statement

http://img241.imageshack.us/img241/7753/scan0001io9.th.jpg [Broken]

Homework Equations

The Attempt at a Solution

i completed the first part fine- knowing the function makes a u shape with min point being 0 and max being 1/2 at both +/- 1. i can't see how using the mean value theorem has much to do with part b) or how part a) helps.

 

[Defennder]

For the mean value theorem, substitute f(c) for f'(c) in the statement, and let b=1 and a=-1. What do you observe? Also take note that by part (a), you have already found the maximum and minimum values of f(c) in the interval [-1,1].

 

[wombat4000]

well from mvt f'(c)=0 which means there is a max/min between a and b - i know its a min, at (0,0). I can't see what any of this had to do with the integral or the inequality?

 

[Defennder]

MVT doesn't say that f'(c) = 0. See here:
http://en.wikipedia.org/wiki/Mean_value_theorem.

Write out that expression, let b=1, a=-1. And then you should realize that f(c) is constrained less than the max value and greater than the min value which you have found in (a).

 

[wombat4000]

but f(a) = f(b) so f'(c)=0?

 

[Defennder]

What does f here refer to? Does it mean the f(x) in the question? If so, then what about F(x)? You didn't make use of that at all, which is required to solve the problem.

 

[exk]

if you have to use an integral perhaps you should use the mvt for integrals?

 

[Defennder]

No, that isn't required. The only thing required is this: $$f'(c) = \frac{f(b)-f(a)}{b-a}$$. And the values of min and max as stated above.

 

[wombat4000]

does this mean i have to integrate f(x)? if so how? i have tried substitution.

 

[Defennder]

No you don't have to. In fact all you have to do is to consider what should f' and f be in the statement of the MVT in relation to this problem. You have to relate these 2 functions to F(x) and f(x) in the question.

 

[Gib Z]

I might be easier to just integrate each part of the inequality produced in a) between -1 and 1.

 

[HallsofIvy]

It certainly is not necessary to actually integrate f.

The mean value theorem says that there exist c between a and b such that
$$\frac{F(b)- F(a)}{b- a}= F'(c)$$

You know that F'(x)= f(x) and from (a) you know the largest and smallest values of f on [-1, 1]. Put those in for F'(c).