Find all numbers c that satisfy the mean value theorem, f(x)=e^2x, [0,3]

Question

anonymous · Answer

mean value theorem states, that if a function is continuous and differentiable on a given interval [w,z], then there exists a value v, where the derivative of the value of the derivative of the function at v is equal to $$f'(v) = (f(z) - f(w))/(z-w)$$

anonymous · Answer

all right, so if $$f(x)=e^{2x}$$ differentiable on [0,3]?

anonymous · Answer

i'd say yes, we can graph the function and it is continuous and differentiable

anonymous · Answer

so, $$f'(x) = 2 * e^{2x}$$

anonymous · Answer

so evaluate f(3) and f(0)

anonymous · Answer

solve for the value c, such that $$f'(c) = 2*e^{2x} = (f(3) - f(0))/(3 - 0)$$

anonymous · Answer

$$f(3) = e^{6}, f(0) = 1$$

anonymous · Answer

so $$(e^6 - 1) / 3 = 2*e^{2*c}$$  solve for c!