# find all values of c in the open interval (a, b) such that f '(c) = 0 If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0 f (x) = ( x / 2 ) − sin (πx / 6), [−1, 0] answers to four decimal places. c =

You need to evaluate the derivative of the given function, such that:

`f'(x) = (x/2 − sin(x*pi/6))' => f'(x) = 1/2 - pi/6*cos(x*pi/6)`

You need to find the value of `c in [-1,0]`  such that `f'(c) = 0` , hence, substituting c for x in equation `f'(x) = 0`  yields:

`1/2 - pi/6*cos(c*pi/6) = 0 =>- pi/6*cos(c*pi/6) = -1/2`

`pi/3*cos(c*pi/6) = 1 => cos(c*pi/6) = 1/(pi/3)`

`cos(c*pi/6) = 3/pi => cos(c*pi/6) = 0.9549 => c*pi/6 = +-cos^(-1)(0.9549)`

`c*pi/6 = +-0.3014 => c = +-6*0.3014/pi => c = +-0.5756`

Since the value `c = 0.5756`  is not in interval `[-1,0], ` hence, the only valid solution to the equation `f'(c) = 0`  is `c = -0.5756` .

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