Question

Find the number c that satisfies the conclusion of the Mean Value Theorem.

f(x)=e^(-3x)
[0,2] Find the number c that satisfies the conclusion of the Mean Value Theorem.

f(x)=e^(-3x)
[0,2] @Mathematics

Answer 1

solve
\[\frac{e^{-6}-e^0}{2}=-3e^{-3c}\] for c

Answer 2

Thanks for the reply! I have to head back to the dorm, will look at it in about 30minutes. Thanks a bunch satellite! REALLY appreciate it!

Answer 3

ok i will write out the first few steps, you can check back later

Answer 4

first of all you are using
\[\frac{f(b)-f(a)}{b-a}=f'(c)\] with
\[f(x)=e^{-3x}, f'(x)=-3e^{-3x}, a = 0, b = 2 \] giving
\[\frac{e^{-6}-e^0}{2-0}=\frac{e^{-6}-1}{2}\]
\[=\frac{1-e^6}{2e^6}\]
so you need to solve
\[\frac{1-e^6}{2e^6}=-3e^{-3c}\] or
\[\frac{1-e^6}{2e^6}=-\frac{3}{e^{3c}}\] and it is algebra from here on in.

Answer 5

get
\[\frac{2e^6}{1-e^6}=-\frac{e^{3c}}{3}\]
\[\frac{6e^6}{e^6-1}=e^{3c}\]
\[3c=\ln(6)+\ln(e^6)-\ln(e^6-1)\]
\[c=\frac{1}{3}(\ln(6)+6-\ln(e^6-1))\]

Answer 6

how did you get from (e^-6 -1 )/2 to 1-e^-6/2e^6

Answer 7

multiplied top and bottom by
\[e^6\] to get rid of the compound fraction

Answer 8

it should be
\[\frac{e^{-6}-1}{2}=\frac{1-e^6}{2e^6}\] the negative exponents annoyed me, because it is really a compound fraction. you can probably work with them but i like to see what is really going on to solve for c

Answer 9

Got it. Thanks bud!

Answer 10

yw