If f(x) = cos x for 0 le x le frac{pi}{2},the | vedclass

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(C) The Mean Value Theorem states that there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \cos x$ on $[0,

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$\sin^{-1}\left(\frac{2}{\pi}\right)$

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(C) The Mean Value Theorem states that there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \cos x$ on $[0, \frac{\pi}{2}]$,we have $a = 0$ and $b = \frac{\pi}{2}$.$f(a) = f(0) = \cos(0) = 1$.$f(b) = f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$.Now,calculate the slope: $\frac{f(b) - f(a)}{b - a} = \frac{0 - 1}{\frac{\pi}{2} - 0} = -\frac{2}{\pi}$.Since $f'(x) = -\sin x$,we set $f'(c) = -\sin c = -\frac{2}{\pi}$.Therefore,$\sin c = \frac{2}{\pi}$,which implies $c = \sin^{-1}\left(\frac{2}{\pi}\right)$.

If $f(x) = \cos x$ for $0 \le x \le \frac{\pi}{2}$,then the real number $c$ of the Mean Value Theorem is:

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