1. ## Rolle's Theorem

hey guys and gals! i'd greatly appreciate if you helped me with some insight on these problems.

1)
Determine whether the Mean Value Theorem is applicable and if it is find c.

$f(x) = e^{-2x}\;,\;[0,3]$
f is continuous on [0,3] because there are no breaks, holes, asymptotes or gaps. f is differentiable on (0,3) because there are no discontinuities previously mentioned or sharp turns.

So there exists a $f\,'\,(c) = \frac{e^{-6} - 1}{3}$

So now I found the derivative to find what c value equals f ' (c).

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

So I set f ' (c) and f ' (x) equal to each other: $-2e^{-2x} = \frac{e^{-6} - 1}{3}$

Then I put the $e^{-2x}$ on the bottom: $\frac{-2}{e^{2x}} = \frac{e^{-6} - 1 }{3}$

I cross multiplied: $-6 = e^{2x - 6} - e^{2x}$

Then I thought take ln of both sides, but ln(-6) gives an error. Why doesn't this work?

2. Originally Posted by Jonboy
hey guys and gals! i'd greatly appreciate if you helped me with some insight on these problems.

1) f is continuous on [0,3] because there are no breaks, holes, asymptotes or gaps. f is differentiable on (0,3) because there are no discontinuities previously mentioned or sharp turns.

So there exists a $f\,'\,(c) = \frac{e^{-6} - 1}{3}$

So now I found the derivative to find what c value equals f ' (c).

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

So I set f ' (c) and f ' (x) equal to each other: $-2e^{-2x} = \frac{e^{-6} - 1}{3}$

Then I put the $e^{-2x}$ on the bottom: $\frac{-2}{e^{2x}} = \frac{e^{-6} - 1 }{3}$

I cross multiplied: $-6 = e^{2x - 6} - e^{2x}$

Then I thought take ln of both sides, but ln(-6) gives an error. Why doesn't this work?
$\frac{e^{-6} - 1}{3}$ is just a number, call it $N$.

Now solve $-2 e^{-2c} = N$ for $c$:

$c = - \frac{1}{2} \ln \left( - \frac{N}{2} \right)$

$= - \frac{1}{2} \ln \left( \frac{e^{-6} - 1}{-6} \right) = - \frac{1}{2} \ln \left( \frac{1 - e^{-6}}{6} \right)$.

3. Originally Posted by Jonboy
$f\,'\,(x) = e^{-2x}(-2) = -2e^{-2x}$

So...I cross multiplied: $-6 = e^{2x - 6} - e^{2x}$

Then I thought take ln of both sides, but ln(-6) gives an error. Why doesn't this work?
You get an error because logs are defined only for positive inputs.

To get something you can work with, try multiplying through by -1:

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

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

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

Then take the log of either side:

. . . . . $-2x\, =\, \ln\left(\frac{1\, -\, e^{-6}}{6}\right)$

. . . . . $x\, =\, \frac{\ln(1\, -\, e^{-6})\, -\, \ln(6)}{-2}$

. . . . . . . $=\, \frac{\ln(6)\, -\, ln(1\, -\, e^{-6})}{2}$

Hope that helps!