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)=x^{2} and the intervals are [-2, 1]

**Answer is: f'(-1/2)=-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)=c^{2}, 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)=x^{2/3 }with intervals [0, 1]

**Answer is: f'(8/27)****=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 :)

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.

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)=x^{2/3} with intervals [0, 1]

Answer is: f'(8/27)=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 c^{2/3}

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

Re: Mean Value Theorem on indicated intervals

Also just wondering, if I have something like x(x^{2}-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(c^{2}-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?

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:

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

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

I would add to your rep, but I gotta spread it xD.