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hey, how's it going?

Good thank you

f'(x) = -2e^(-2x) Sorry Interval [0,3]
a @ 0 = 1
b @ 3 = e^-6

so what are you stuck on?

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)]

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!