Question

Find the number c that satisfies the conclusion of the Mean Value Theorem.
f(x) = e^-5x
[0, 1]
so I did (e^-5(1))-(e^5(0))/1-0
that gave me (e^-5)-1
but when I plug that in it won't work, so I figured that since the answer ends up being -.9933 and that the interval is [1,0] it would be DNE, but that isn't right either.
help?

Answer 1

\[f'(c)=\frac{f(b)-f(a)}{b-a}\]

Answer 2

\[-5e^{-5c}=\frac{e^{-5(1)}-e^{-5(0)}}{1-0}\]

Answer 3

\[-5e^{-5c}=\frac{e^{-5}-1}{1}\]

Answer 4

\[-5e^{-5c}=e^{-5}-1\]

Answer 5

solve for c

Answer 6

mmm could you explain why you set it equal to -5e^-5c?

Answer 7

its the formula i wrote above
\[e^{-5c}=\frac{e^{-5}-1}{-5}\]

Answer 8

\[f'(x)=(-5)e^{-5x} =>f'(c)=-5e^{-5c}\]

Answer 9

ahh okay thank you :)

Answer 10

\[\ln(e^{-5x})=\ln(\frac{e^{-5}-1}{-5}) => -5x=\ln(\frac{e^{-5}-1}{-5})\]

Answer 11

i meant that to be c not x (but whatever)
\[c=\frac{-1}{5} \cdot \ln(\frac{e^{-5}-1}{-5})\]
and this c is in btw 0 and 1