`f(x) = sin(x), [0,pi]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.
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"it is `pi/2` ", some bug occured.
Yes, it can. The function f is contionuous on ` [0, pi ]` and is differentiable on `(0, pi )`, which are all the requirements for the Mean Value Theorem.
Therefore there is at least one point c on `(0, pi)` such that
`f'(c) = (f(pi)-f(0))/(pi-0) = (0-0)/pi = 0.`
Let's find all these points:
f'(c) = cos(c) = 0.
`c = pi/2 + k*pi` where k is integer, there is only one such point on` (0, pi)`, it is .
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