The Mean Value Theorem (MVT) states that if a function f is continuous on the 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).
Your function f(x)=x³-3x² 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.
Since f'(x)=3x²-6x, we have f'(c)=3c²-6c=0, which has only one solution in (0,3), namely c=2 (since the other solution, c=0, is not in the open interval).
Sun K.
01/07/14