Use the graph of f(x)=x^2/(x^2-4) to determine on which of the following intervals Rolle’s Theorem applies.
A. [0, 3]
B. [-3, 3]
C. [-3/2, 3/2]
D. [-2, 2]
E. none of these
I know what the Rolle's Theorem is but I'm unsure on how you know if the function is continuous and differentiable. The domain I think is all except -2 and 2 so how would that affect the differentiable side. I have eliminated choice A because f(0) and f(3) do not equal each other.
I'm just confused.
Since f(x) is not defined for x = ±2, any interval which includes either of those values will not satisfy the theorem.
So, A,B,D are all out
The only possible choice is C, which is ok, because f(-3/2) = f(3/2)
To determine on which intervals Rolle's Theorem applies, we need to check if the conditions of Rolle's Theorem are satisfied. Rolle's Theorem states that for a function f(x) to have a point c in the interval (a, b) where f'(c) = 0, three conditions must be met:
1. Continuity: The function f(x) must be continuous on the closed interval [a, b]. This means there should be no holes, jumps, or asymptotes in the graph of f(x) within the interval.
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b). This means that the derivative f'(x) must exist and be defined for all x in the interval.
3. Equal function values: The function f(x) must have equal function values at the endpoints of the interval, i.e., f(a) = f(b).
In this case, the function is given as f(x) = x^2/(x^2 - 4).
To determine if the function is continuous, we need to check if there are any holes, jumps, or asymptotes in the graph. The domain of f(x) is all real numbers except x = 2 and x = -2, which means the graph has vertical asymptotes at x = 2 and x = -2. Since the graph has vertical asymptotes, it is not continuous at x = 2 and x = -2.
Therefore, we can eliminate choice D ([-2, 2]) as a possible interval.
Now, let's check the remaining intervals:
A. [0, 3]: As you mentioned, f(0) and f(3) do not equal each other, so this interval does not satisfy the third condition of Rolle's Theorem (equal function values at the endpoints).
B. [-3, 3]: We need to check if the function is continuous and differentiable on this interval. The function is undefined at x = 2 and x = -2, but both of these points are outside the interval [-3, 3]. Therefore, the function is continuous on [-3, 3]. To check differentiability, we need to make sure that f'(x) exists and is defined for all x in (-3, 3). The derivative of f(x) is f'(x) = (2x(x^2 + 4))/(x^2 - 4)^2, which is defined for all x except x = 2 and x = -2. Since both x = 2 and x = -2 are outside the interval [-3, 3], the function is differentiable on (-3, 3).
C. [-3/2, 3/2]: Similar to choice B, we need to check continuity and differentiability on this interval. The function is undefined at x = 2 and x = -2, but both points are outside the interval [-3/2, 3/2]. Therefore, the function is continuous. Additionally, the derivative f'(x) is defined for all x except x = 2 and x = -2, which are outside the interval of interest. Hence, the function is differentiable on (-3/2, 3/2).
Both B. [-3, 3] and C. [-3/2, 3/2] satisfy the conditions of Rolle's Theorem. Therefore, the answer is either B or C.
Hope this clarifies your confusion!