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

determine whether the mean value theorem can be applied to ( f ) on the closed interval (a, b). (select all that apply.)\n\n( f(x)=4 sin (x), quad0, pi )\n\n( square ) yes, the mean value theorem can be applied.\n\n( square ) no, ( f ) is not continuous on (a, b).\n\n( square ) no, ( f ) is not differentiable on ( (a, b) ).\n\n( square ) none of the above.\n\nif 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.)\n\n( c= )

determine whether the mean value theorem can be applied to ( f ) on the closed interval (a, b). (select all that apply.)\n\n( f(x)=4 sin (x), quad0, pi )\n\n( square ) yes, the mean value theorem can be applied.\n\n( square ) no, ( f ) is not continuous on (a, b).\n\n( square ) no, ( f ) is not differentiable on ( (a, b) ).\n\n( square ) none of the above.\n\nif 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.)\n\n( c= )

Answer

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})