Just making sure if I got mean value theoroms,

Apply the Mean Value Theorem tofon 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, buthow did they get the (-1/2) in there.Another example is like f(x)=x

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.

^{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

ButI'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