Thread: 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]
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)=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)=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

For the first one, we then want to set:







So we know

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

Originally Posted by MarkFL2

For the first one, we then want to set:







So we know

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!!!

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?

You want to set:









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

Originally Posted by MarkFL2

You want to set:









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

Oh right!
Forgot to take the derivative!
Thanks!
I'm starting to understand the Mean Value Theorom now
I would add to your rep, but I gotta spread it xD.