00:01
Wish to determine where this function satisfies the mean value theorem.
00:06
We know that it does, it does satisfy the mean value theorem because although it would be discontinuous and non -differentiable at one, that's outside the interval.
00:18
So it is equal to, or it is differentiable on the open interval and it is continuous on the closed.
00:27
By definition, mean value of the theorem, says the derivative at some value inside the interval will equal the average rate of change from the beginning to end f of b minus f of a over b minus a now i'm going to take the derivative at x and just realize that that x is the c value the derivative is one well derivative of two is zero but the derivative of three over x minus one could be we could rewrite the function is three times the quantity x minus 1 to the negative 1st.
01:07
The derivative would use the chain rule.
01:09
We'd bring the power out front and subtract one from the old power, so make that x minus 1 to the negative second.
01:18
And then we'd multiply by the derivative inside, but that's 1.
01:22
The function at 7 would be 9 plus 3 over 4.
01:28
So that's 9 and 3 4ths, or 39 4ths.
01:32
If we fill in two, that's going to be two plus two plus three.
01:39
That's seven, and that's all over seven minus two.
01:44
So we have one minus three quantity...