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hey, how's it going?
I know the answer to this problem, but have gotten stuck when plugging in the values of a,b into the derivative
Good thank you
f'(x) = -2e^(-2x) Sorry Interval [0,3] a @ 0 = 1 b @ 3 = e^-6
so what are you stuck on?
is this definitely the question as written, it doesn't look like the answer is going to be very "pretty"
f(x)=e^(-2x) Find all numbers that satisfy the conclusion of the Mean Value Theorem
on the closed interval [0,3]
ok so what are you stuck on
Getting the answer of -1/2ln[1/6(1-e^-6)]
do you know the formula of the mean value theorem? maybe you can tell me what part you're having trouble with
f'(c)=f(b)-f(a)/b-a or f(b)-f(a) = f'(c)(b-a)
do you want to tell me the steps you took and maybe i can see where you made a mistake?
I am not clear on how to incorporate the deriviative which is -2e^-2x using my a and b into this...
walking myself in circles
c is what you want to find
you have b and a and f(b) and f(a)
so f '(c) = -2e^(-2c) = you have the formula
the theorem means that there is an x-value c somewhere in the interval that makes the formula true
ok. so then would it e^(-6)-1=-2e^(-2c)(e^-6-1)
this being the average rate of change between these points, right
sorry last part (3-0)
looks right i think with your correction
Sometimes it helps to talk it out. Thanks a million!
did you end up getting the right answer?
Working through it now. YES!