# How do I find the numbers c that satisfy the Mean Value Theorem for f(x)=e^(-2x) on the interval [0,3] ?

Sep 12, 2014

The Mean Value Theorem guarantees that there exists a number $c$ in $\left(0 , 3\right)$ such that $f ' \left(c\right) = \frac{f \left(3\right) - f \left(0\right)}{3 - 0}$. The actual value is
$c = \ln \sqrt{\frac{6 {e}^{6}}{{e}^{6} - 1}} \approx 0.9$.

Let us find such $c$.

By Chain Rule,
$f ' \left(x\right) = - 2 {e}^{- 2 x}$

So, the left-hand side is
$f ' \left(c\right) = - 2 {e}^{- 2 c}$

Let us find the right-hand side.
$\frac{f \left(3\right) - f \left(0\right)}{3 - 0} = \frac{{e}^{- 6} - 1}{3}$
by multiplying the numerator and the denominator by ${e}^{6}$,
$= \frac{1 - {e}^{6}}{3 {e}^{6}}$

By setting the left-hand side and the right-hand side equal to each other,
$- 2 {e}^{- 2 c} = = \frac{1 - {e}^{6}}{3 {e}^{6}}$

By dividing by -2,
${e}^{- 2 c} = \frac{{e}^{6} - 1}{3 {e}^{6}}$

By taking the natural log,
$- 2 c = \ln \left(\frac{{e}^{6} - 1}{6 {e}^{6}}\right)$

By dividing by -2,
$c = - \frac{1}{2} \ln \left(\frac{{e}^{6} - 1}{6 {e}^{6}}\right)$

by the log property $r \ln x = \ln {x}^{r}$,
$= \ln {\left(\frac{{e}^{6} - 1}{6 {e}^{6}}\right)}^{- \frac{1}{2}}$

by simplifying a bit further,
$= \ln \sqrt{\frac{6 {e}^{6}}{{e}^{6} - 1}} \approx 0.9$