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)=x2/3 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