determine whether rolles theorem can be applied to ( f ) on the closed interval (a, b). (select all that…

Explanation:

Step1: Check continuity

The function (y = \cos(x)) is a basic trigonometric function. The function (f(x)=7\cos(\pi x)) is a composition of a constant - multiple of a cosine function. Since (y = \cos(u)) is continuous for all (u\in R) and (u = \pi x) is a polynomial (continuous for all (x\in R)), (f(x)) is continuous on the closed interval ([0,2]).

Step2: Check differentiability

The derivative of (f(x)) using the chain rule: if (y = 7\cos(u)) and (u=\pi x), then (y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}). We know that (\frac{d}{du}(\cos(u))=-\sin(u)) and (\frac{d}{dx}(\pi x)=\pi). So (f^\prime(x)=- 7\pi\sin(\pi x)). The function (y = \sin(x)) is continuous for all (x\in R), so (f(x)) is differentiable on the open interval ((0,2)).

Step3: Check (f(a)=f(b))

Calculate (f(0)) and (f(2)). When (x = 0), (f(0)=7\cos(0)=7\times1 = 7). When (x = 2), (f(2)=7\cos(2\pi)=7\times1 = 7). So (f(0)=f(2)).

Step4: Find (c)

Since Rolle's Theorem can be applied ((f(x)) is continuous on ([0,2]), differentiable on ((0,2)) and (f(0)=f(2))), we set (f^\prime(c)=0). We have (f^\prime(x)=-7\pi\sin(\pi x)), and (f^\prime(c)=-7\pi\sin(\pi c)=0). (\sin(\pi c)=0), then (\pi c = k\pi), where (k\in Z). So (c = k). For (c\in(0,2)), when (k = 1), (c = 1).

Answer:

Yes, Rolle's Theorem can be applied. (c = 1)