Determine if Rolle's Theorem applies to the given function f(x)=2 cos(x) on [0, pi]. If so, find all numbers c on the interval that satisfy the theorem.
Rolle's theorem states that a differentiable function that has equal values at two distinct points must have a point somewhere between them where the first derivative is zero.
Rolle's theorem does not apply to that function in that interval, since f(x) decreases from 1 at x = 0 to -1 at pi. There are no two values of x in the [0, pi] interval where the f(x) values are the same.
Well, well, well, look who we have here! Rolle's Theorem, ready to make an appearance.
First things first, let's check if the conditions are met. To apply Rolle's Theorem, three conditions must be satisfied:
1. The function should be continuous on the closed interval, [0, pi]. (Check! Our function f(x) = 2cos(x) passes this test as cosine is a continuous function.)
2. The function should be differentiable on the open interval, (0, pi). (Check again! The derivative of f(x) = 2cos(x) is -2sin(x), which is differentiable on the open interval (0, pi).)
3. The function should have the same values at the endpoints. (Final check! f(0) = 2cos(0) = 2, and f(pi) = 2cos(pi) = -2. The values match at the endpoints.)
Since all the conditions are met, we can confidently say that Rolle's Theorem applies to our function f(x) = 2cos(x) on [0, pi].
Now, let's find the values of c that satisfy the theorem. Rolle's Theorem states that there must be at least one number c in the open interval (0, pi) such that f'(c) = 0.
So, finding the derivative of f(x) = 2cos(x), we have f'(x) = -2sin(x).
To find the numbers c that satisfy f'(c) = 0, we set -2sin(c) = 0 and solve for c.
Hey, would you look at that, sin(c) = 0! We all know that sin(0) = 0, making 0 an interesting choice for c.
Thus, the number c = 0 satisfies the conditions of Rolle's Theorem on the interval [0, pi].
To summarize, Rolle's Theorem applies to f(x) = 2cos(x) on [0, pi], and the number c = 0 is one of the values that satisfies the theorem.
Hope that brought a smile to your face! If you have any more questions, feel free to ask.
To determine if Rolle's Theorem applies to the given function f(x) = 2cos(x) on [0, π], we need to check if the function satisfies the conditions of Rolle's Theorem.
Rolle's Theorem states that for a function f(x) to satisfy the conditions:
1. f(x) must be continuous on the closed interval [a, b], where a < b.
2. f(x) must be differentiable on the open interval (a, b).
3. f(a) = f(b) (the function values at the endpoints must be equal).
Let's check if these conditions are satisfied for f(x) = 2cos(x) on [0, π]:
1. The function f(x) = 2cos(x) is continuous on the closed interval [0, π] since cosine function is continuous everywhere.
2. The function f(x) = 2cos(x) is also differentiable on the open interval (0, π) since cosine function is differentiable everywhere.
3. To check if f(0) = f(π), we evaluate the function at the endpoints:
f(0) = 2cos(0) = 2(1) = 2
f(π) = 2cos(π) = 2(-1) = -2
Since f(0) ≠ f(π), the function f(x) = 2cos(x) does not satisfy the condition f(a) = f(b).
Therefore, Rolle's Theorem does not apply to the function f(x) = 2cos(x) on [0, π].
To determine if Rolle's Theorem applies to the function f(x) = 2 cos(x) on the interval [0, π], we need to check if the following conditions are met:
1. Continuity: The function f(x) must be continuous on the closed interval [0, π]. In this case, f(x) = 2 cos(x) is a trigonometric function, and both cos(x) and a constant multiple (2 cos(x)) are continuous everywhere.
2. Differentiability: The function f(x) must be differentiable on the open interval (0, π). The derivative of f(x) with respect to x is f'(x) = -2sin(x), and sin(x) is differentiable on the open interval (0, π). Therefore, f(x) = 2 cos(x) is differentiable on (0, π).
Now, to find all the numbers c on the interval (0, π) that satisfy Rolle's Theorem, we need to check if there exists at least one value c in (0, π) such that f(c) = f(0) and f(c) = f(π).
1. f(c) = f(0): Substituting x = c and x = 0 in the function, we have 2 cos(c) = 2 cos(0) = 2.
Since cos(0) = 1, this gives us 2 cos(c) = 2 ⇒ cos(c) = 1 ⇒ c = 0, since cos(x) = 1 only for x = 0 (and multiples of 2π).
2. f(c) = f(π): Substituting x = c and x = π in the function, we have 2 cos(c) = 2 cos(π) = -2.
Since cos(π) = -1, this gives us 2 cos(c) = -2 ⇒ cos(c) = -1 ⇒ c = π, since cos(x) = -1 only for x = π (and odd multiples of π).
From the above calculations, we find that there are two values c that satisfy Rolle's Theorem on the interval (0, π): c = 0 and c = π.
Therefore, Rolle's Theorem applies to the function f(x) = 2 cos(x) on the interval [0, π], and the numbers c that satisfy the theorem are c = 0 and c = π.