1. Verify that the function satisfies the mean value theorem, then find all numbers c that satisfy the the conclusion of the mvt f(x)=e^(-2x) on the interval [0,3] 2. f'(c)=[f(b)-f(a)]/[b-a] 3. 1. f(x) is a composition of continuous functions, so f(x) is continuous on[0,3]. 2. f(x) is a composition of differentiable functions, so it is differentiable on (0,3). so, f(b)-f(a)=f'(c)(b-a) e^-6-e^0=(-2*e^-2c)(3) Then take the ln of both sides... -6-0=ln(-6*e^-2c) -6=ln(-6)+2c Solve for c..? c=[-6-ln(-6)]/[-2] Can't have ln of a negative..(well without imaginary numbers). What's gone wrong? I'm stumped.