Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(x)=0.
f(x) = x^(2/3) - 1 [-8,8]
I plugged in both values and found out that they both equal 4 so I thought Rolle's theorem applied.
However, the answer is that this function is not differentiable at 0.
Can someone explain to me why it is not differentiable even though f(a)=f(b).
correct. f'(x) = 2/3∛x
which is not defined at x=0.
so, the requirements of the theorem are not met. There is, in fact no value of x where f'(x) = 0.
To determine whether Rolle's Theorem can be applied to a function on a closed interval [a, b], there are three conditions that need to be satisfied:
1. Continuity: The function f(x) must be continuous on the closed interval [a, b].
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b).
3. Endpoint equality: The function f(a) must be equal to f(b).
In the given function f(x) = x^(2/3) - 1 on the closed interval [-8, 8], let's examine the three conditions:
1. Continuity: The function f(x) is continuous on the closed interval [-8, 8] since it is a polynomial-like function and does not have any singular points or jumps.
2. Differentiability: To check differentiability, we need to verify whether the derivative f'(x) exists for all points within the open interval (-8, 8). Let's differentiate f(x) using the power rule:
f'(x) = (2/3)x^(-1/3)
However, notice that at x = 0, the derivative f'(x) becomes undefined since x^(-1/3) is not defined for x = 0. Therefore, the function is not differentiable at x = 0, which means it does not satisfy the differentiability condition for Rolle's Theorem.
3. Endpoint equality: In this case, f(-8) = (-8)^(2/3) - 1 = 3, and f(8) = (8)^(2/3) - 1 = 3. Thus, f(-8) is equal to f(8).
Since the function fails the differentiability condition at x = 0, Rolle's Theorem cannot be applied to find values of c such that f'(c) = 0 in the open interval (-8, 8).