For the function f(x) = log(sin x) in the int | vedclass

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(B) Given function is $f(x) = \log(\sin x)$.According to Lagrange's Mean Value Theorem $(LMVT)$,there exists a point $c \in (a, b)$ such that $f'(c) =

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$\frac{\pi}{2}$

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(B) Given function is $f(x) = \log(\sin x)$.According to Lagrange's Mean Value Theorem $(LMVT)$,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = \frac{\pi}{6}$ and $b = \frac{5\pi}{6}$.$f(a) = \log(\sin \frac{\pi}{6}) = \log(\frac{1}{2})$ and $f(b) = \log(\sin \frac{5\pi}{6}) = \log(\frac{1}{2})$.Thus,$f'(c) = \frac{\log(1/2) - \log(1/2)}{\frac{5\pi}{6} - \frac{\pi}{6}} = 0$.Now,$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.So,$\cot c = 0$.Since $c \in (\frac{\pi}{6}, \frac{5\pi}{6})$,the value of $c$ is $\frac{\pi}{2}$.

For the function $f(x) = \log(\sin x)$ in the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$,what is the value of $c$ according to Lagrange's Mean Value Theorem?

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