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AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1mean 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 \[f'(v) = (f(z)  f(w))/(zw)\]

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1all right, so if \[f(x)=e^{2x}\] differentiable on [0,3]?

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1i'd say yes, we can graph the function and it is continuous and differentiable

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1so, \[f'(x) = 2 * e^{2x}\]

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1so evaluate f(3) and f(0)

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1solve for the value c, such that \[f'(c) = 2*e^{2x} = (f(3)  f(0))/(3  0)\]

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1\[f(3) = e^{6}, f(0) = 1\]

AntiMatter
 3 years ago
Best ResponseYou've already chosen the best response.1so \[(e^6  1) / 3 = 2*e^{2*c}\] solve for c!
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