Question

Let f be the function given by f(x)=x³−3x². What are all values of c that satisfy the conclusion of the Mean Value Theorem of differential calculus on the closed interval [0,3]?

Answer 1

Answer:

The values are c = 2

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 one solution in (0, 3), namely c = 2 ( since the other solution, c = 0, is not in the open interval).