Question

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

( f(x)=4 sin (x), quad0, pi )

( square ) yes, the mean value theorem can be applied.

( square ) no, ( f ) is not continuous on (a, b).

( square ) no, ( f ) is not differentiable on ( (a, b) ).

( square ) none of the above.

if the mean value theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=\frac{f(b)-f(a)}{b - a} ). (enter your answers as a comma-separated list. if the mean value theorem cannot be applied, enter na.)

( c= )

Explanation:

Step1: Check continuity and differentiability

The function \(y = \sin(x)\) is continuous everywhere, so \(f(x)=4\sin(x)\) is continuous on \([0,\pi]\). The derivative \(f'(x) = 4\cos(x)\) exists for all \(x\in(0,\pi)\), so \(f(x)\) is differentiable on \((0,\pi)\). Thus, the Mean - Value Theorem can be applied.

Step2: Calculate \(f(a)\) and \(f(b)\)

Given \(a = 0\), \(b=\pi\), \(f(a)=4\sin(0)=0\), \(f(b)=4\sin(\pi)=0\).

Step3: Use the Mean - Value Theorem formula

The Mean - Value Theorem states that \(f'(c)=\frac{f(b)-f(a)}{b - a}\). Since \(\frac{f(b)-f(a)}{b - a}=\frac{0 - 0}{\pi-0}=0\), and \(f'(x)=4\cos(x)\), we set \(4\cos(c)=0\).

Step4: Solve for \(c\)

Dividing both sides of \(4\cos(c)=0\) by \(4\) gives \(\cos(c)=0\). On the interval \((0,\pi)\), \(c=\frac{\pi}{2}\) (because \(\cos(x) = 0\) when \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\), and for \(x\in(0,\pi)\), \(k = 0\)).

Answer:

Yes, the Mean Value Theorem can be applied. \(c=\frac{\pi}{2}\)