Consider the function f(x) = 3-3 x^{2/3} on the interval [ -1 , 1 ].
Which of the three hypotheses of Rolle's Theorem fails for this function on the inverval?
(a) f(x) is continuous on [-1,1].
(b) f(x) is differentiable on (-1,1).
(c) f(-1)=f(1).
Answer:( a, b, or c )
dy/dx = -3*(2/3)x^-(2/3) =-2/x^(2/3)
which is undefined at x = 0
To determine which of the three hypotheses of Rolle's Theorem fails for the function f(x) = 3-3 x^(2/3) on the interval [-1, 1], let's examine each hypothesis one by one.
(a) The first hypothesis states that the function should be continuous on the closed interval [-1, 1]. In this case, the function f(x) = 3-3 x^(2/3) is continuous on the interval [-1, 1]. This means hypothesis (a) is satisfied.
(b) The second hypothesis states that the function should be differentiable on the open interval (-1, 1). To check differentiability, we need to compute the derivative of f(x). Applying the power rule, we have:
f'(x) = (2/3) * (-3) * x^(-1/3) = -2x^(-1/3)
The derivative exists and is defined for all x ≠ 0, which means that the function is differentiable everywhere on the open interval (-1, 1). Thus, hypothesis (b) is also satisfied.
(c) The third hypothesis states that the function should have the same value at the endpoints of the interval, i.e., f(-1) = f(1). Evaluating the function at these points, we have:
f(-1) = 3-3 * (-1)^(2/3) = 3-3 = 0
f(1) = 3-3 * (1)^(2/3) = 3-3 = 0
Here, we can see that f(-1) and f(1) both equal to 0, indicating that the values at the endpoints are the same. Therefore, hypothesis (c) is satisfied as well.
In conclusion, none of the three hypotheses of Rolle's Theorem fails for the function f(x) = 3-3 x^(2/3) on the interval [-1, 1]. Therefore, the answer is "None of the above (a, b, or c)".