f(x) = x^3 - 2x^2 + x
f'(x) = 3x^2 - 4x + 1
so, where is f' = 0?
The function defined below satisfies Rolle's Theorem on the given interval. Find the value of c in the interval (0,1) where f'(c)=0.
f(x) = x3 - 2x2 + x, [0, 1]
Round your answer to two decimal places.
1 answer