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**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 $\displaystyle f\,'\,(c) = \frac{e^{-6} - 1}{3}$

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

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

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

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

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

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