Why would fZ(x) =2x^3 on [-1,1] fail to satisfy the conditions of the Mean Value therorm?
It is continuous and differential. So what is wrong with it I can't figure it out.
I agree with you. The theorem should apply for that "fZ" function over that interval.
Is the Z after f supposed to be there?
The notation "fZ(x)" seems to be a typo or an error because the standard notation for a function is simply "f(x)" without the "Z" subscript. However, regardless of the notation used, we can analyze the function f(x) = 2x^3 on the interval [-1, 1] to determine if it satisfies the conditions of the Mean Value Theorem.
The Mean Value Theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In our case, the function f(x) = 2x^3 is indeed continuous on the closed interval [-1, 1] and differentiable on the open interval (-1, 1). Therefore, it seems to satisfy the conditions necessary for the Mean Value Theorem to be applicable.
However, there might be a misunderstanding or oversight in the original question or context. It is possible that there are additional conditions or specific criteria that the function f(x) should satisfy in order to be able to apply the Mean Value Theorem. Without further information, it is difficult to identify any specific reason why the function f(x) = 2x^3 would fail to satisfy the conditions of the Mean Value Theorem.