# Mean Value Theorem on indicated intervals

• Nov 27th 2012, 07:48 PM
Chaim
Mean Value Theorem on indicated intervals
Just making sure if I got mean value theoroms,
Apply the Mean Value Theorem to f on the indicated interval, and find all values of c in the interval (a, b) such that: f'(c)=(f(b)-f(a))/(b-a)

So if I have something like f(x)=x2 and the intervals are [-2, 1]
I would plug in the intervals first into the equation right?
So f(a)=f(-2)=4 and f(b)=f(1)=1
Then I just plugged in what I know back into that equation up there, which turns out to be (1-4)/(1+2)=(-3/3) = -1
But I'm a bit confused, I got the -1 thing down, but how did they get the (-1/2) in there.
So I thought it was the value of 'c', so I did f(c)=c2, then did f'(c), which is 2c=-1
So c=-1/2, I think, if I did this right, but can someone verify if this is the way to do it.

Another example is like f(x)=x2/3 with intervals [0, 1]
I did the same thing,
So f(a)=f(0)=0 and f(b)=f(1)=1
Then I plugged it back into the equation (1-0)/(1-0)=1
But I'm confused on where they got (8/27)
So I thought f'(c)=(2/3)x-1/3=1, then x-1/3=(3/2), then I got lost on here, since I did both side to the third power, but that was a bit off o.o

I'm not sure if that was a concidence that I get the ending result, but the thing inside the paranthesis inside the answer is what I'm lost at, which is the 'value of c' I believe.

Can someone explain please? Thanks :)
• Nov 27th 2012, 07:54 PM
MarkFL
Re: Mean Value Theorem on indicated intervals
For the first one, we then want to set:

$\displaystyle f'(x)=-1$

$\displaystyle 2x=-1$

$\displaystyle x=-\frac{1}{2}$

So we know $\displaystyle f'\left(-\frac{1}{2} \right)=-1$

See if you can use this technique to get the second one.
• Nov 27th 2012, 08:06 PM
Chaim
Re: Mean Value Theorem on indicated intervals
Quote:

Originally Posted by MarkFL2
For the first one, we then want to set:

$\displaystyle f'(x)=-1$

$\displaystyle 2x=-1$

$\displaystyle x=-\frac{1}{2}$

So we know $\displaystyle f'\left(-\frac{1}{2} \right)=-1$

See if you can use this technique to get the second one.

Ok, so for the 2nd one
f(x)=x2/3 with intervals [0, 1]

Lets see
So first you would use that equation at the top right?
So it's f(b) which is f(1)=1, and f(a)=0
Then the equation is (1-0)/(1-0) = 1

Then you took f(c) to the original equation, which then becomes c2/3
Then the derivative of that is (2/3)c-1/3=1
c-1/3=(3/2)
So I multiplied the exponents both by (-3/1), I think, right? o.o
Which makes it (3/2)(-3/1)

So that means it'll be (3/2)-3, which then turns into (2/3)3, which equals 8/27!!!
Thanks!!! :D
• Nov 27th 2012, 08:22 PM
Chaim
Re: Mean Value Theorem on indicated intervals
Also just wondering, if I have something like x(x2-x-2) intervals [-1, 1]
I got f(-1)=0
f(1)=-2
Then (-2-0)/(1-(-1)) = -2/2 = -1
Then I did f(c)=c(c2-c-2)=-1
Then I got c(c-1)(c-2)=-1
I was a bit confused on this one, where there are more of the same variable, and the end is -1, so it's hard to find out since regularly it would be 0.

So how would I find what c is here?
• Nov 27th 2012, 08:29 PM
MarkFL
Re: Mean Value Theorem on indicated intervals
You want to set:

$\displaystyle f'(x)=-1$

$\displaystyle 3x^2-2x-2=-1$

$\displaystyle 3x^2-2x-1=0$

$\displaystyle (3x+1)(x-1)=0$

Discard the root not within the open interval (-1,1) to get:

$\displaystyle c=-\frac{1}{3}$
• Nov 27th 2012, 08:35 PM
Chaim
Re: Mean Value Theorem on indicated intervals
Quote:

Originally Posted by MarkFL2
You want to set:

$\displaystyle f'(x)=-1$

$\displaystyle 3x^2-2x-2=-1$

$\displaystyle 3x^2-2x-1=0$

$\displaystyle (3x+1)(x-1)=0$

Discard the root not within the open interval (-1,1) to get:

$\displaystyle c=-\frac{1}{3}$

Oh right!
Forgot to take the derivative!
Thanks! :)
I'm starting to understand the Mean Value Theorom now :D