Mean value theorum help?

• Apr 20th 2010, 05:42 PM
dorkymichelle
Mean value theorum help?
Find all numbers c that satisfy the conclusion of the Mean Value Theorem.
so f(x)=e ^ -2x [0,3]
using f(b)-f(a)=f'(c)(b-a)
so that's
f'(x) = 2e^-2x
e^-2(3)-e^-2(0) = 2e^-2x(3-0)
e^-6-1=3e^-2x
now how do i solve for x?
the ^ means exponent,
the latex tags won't turn it into an exponent, how weird...
• Apr 20th 2010, 05:47 PM
pickslides
Quote:

Originally Posted by dorkymichelle
$\displaystyle e^-6-1=3e^-2x$
now how do i solve for x?

$\displaystyle e^{-6}-1=3e^{-2x}$

If so divide both sides by 3, take the natural log of both sides and then divide both sides by -2
• Apr 20th 2010, 05:49 PM
skeeter
Quote:

Originally Posted by dorkymichelle
Find all numbers c that satisfy the conclusion of the Mean Value Theorem.
so $\displaystyle f(x)=e^-2x [0,3]$
using $\displaystyle f(b)-f(a)=f'(c)(b-a)$
so that's
$\displaystyle f'(x) = 2e^-2x$
$\displaystyle e^-2(3)-e^-2(0) = 2e^-2x(3-0)$
$\displaystyle e^-6-1=3e^-2x$
now how do i solve for x?

$\displaystyle f'(x) = -2e^{-2x}$

$\displaystyle \frac{e^{-6} - 1}{3} = -2e^{-2x}$

$\displaystyle \frac{1 - e^{-6}}{6} = e^{-2x}$

log of both sides ... then isolate x