Here's the question you clicked on:
destinc
Find all numbers c that satisfy the mean value theorem, f(x)=e^2x, [0,3]
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 \[f'(v) = (f(z) - f(w))/(z-w)\]
all right, so if \[f(x)=e^{2x}\] differentiable on [0,3]?
i'd say yes, we can graph the function and it is continuous and differentiable
so, \[f'(x) = 2 * e^{2x}\]
so evaluate f(3) and f(0)
solve for the value c, such that \[f'(c) = 2*e^{2x} = (f(3) - f(0))/(3 - 0)\]
\[f(3) = e^{6}, f(0) = 1\]
so \[(e^6 - 1) / 3 = 2*e^{2*c}\] solve for c!