Answer:
The values are <em>c = 2 </em>
Step-by-step explanation:
The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that
f'(c) = (f(b) - f(a)) / (b - a)
Therefore,
f(x) = x³ - 3 x² is a function that is continuous on [0, 3] and differentiable on (0, 3), so there exists at least one c in (0, 3) such that
f'(c) = (f(3) - f(0)) / (3 - 0) = (0 - 0) / 3 = 0
where
f'(x) = 3 x² - 6 x,
f(3) = 0
f(0) = 0
Thus, since f'(0) = 3 c² - 6 c = 0, which only has <em>one solution</em> in (0, 3), namely <em>c = 2</em> ( since the other solution, c = 0, is not in the open interval).
