Final answer:
Yes, Rolle's theorem can be applied to the function f(x) = x^(2/3) - 5 on the closed interval [-27, 27].
Step-by-step explanation:
Yes, Rolle's theorem can be applied to the function f(x) = x2/3 - 5 on the closed interval [-27, 27].
Rolle's theorem states that if a function is continuous on a closed interval and differentiable on an open interval, and the function values at the endpoints of the interval are equal, then there exists at least one point in the open interval where the derivative of the function is zero.
In this case, the function f(x) = x2/3 - 5 is continuous on the closed interval [-27, 27] and differentiable on the open interval (-27, 27). The function values at the endpoints are f(-27) = (-27)2/3 - 5 = 4 and f(27) = 272/3 - 5 = 22. Since f(-27) = f(27), we can apply Rolle's theorem and conclude that there exists at least one point in the open interval (-27, 27) where the derivative of the function is zero.