Answer
On the graph, we can see that the tangent line at the point $x=0.8386$ is parallel to the secant line.
Work Step by Step
$f(x) = e^{-x}$
$f'(x) = -e^{-x}$
1. $f$ is continuous on the interval $[0,2]$
2. $f$ is differentiable on the interval $(0,2)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,2]$
According to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,2)$ such that $f'(c) = \frac{f(2)-f(0)}{2-0}$
Then:
$f'(c) = \frac{f(2)-f(0)}{2-0}$
$-e^{-c} = \frac{e^{-2}-e^{-0}}{2-0}$
$e^{-c} = \frac{1-(1/e^2)}{2}$
$-c = ln\Big(\frac{1-(1/e^2)}{2}\Big)$
$c = -ln\Big(\frac{1-(1/e^2)}{2}\Big)$
$c \approx 0.8386$
On the graph, we can see that the tangent line at the point $x=0.8386$ is parallel to the secant line.