Question

determine whether rolles theorem can be applied to ( f ) on the closed interval (a, b). (select all that apply.)
( f(x)=7 cos (pi x), quad0,2 )
yes, rolles theorem can be applied.
no, because ( f ) is not continuous on the closed interval (a, b).
no, because ( f ) is not differentiable on the open interval ( (a, b) ).
no, because ( f(a)
eq f(b) ).
if rolles theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=0 ). (enter your answers as a comma-separated list. if rolles theorem cannot be applied, enter na.)
( c= )

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. \(c = 1\)