Values of c as per Rolle's theorem for f(x)=s | vedclass

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Question

(C) Given function is $f(x) = \sin x + \cos x + 6$ on the interval $[0, 2\pi]$.According to Rolle's theorem,there exists at least one $c \in (0, 2\pi)

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$\frac{\pi}{4}, \frac{5\pi}{4}$

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(C) Given function is $f(x) = \sin x + \cos x + 6$ on the interval $[0, 2\pi]$.According to Rolle's theorem,there exists at least one $c \in (0, 2\pi)$ such that $f'(c) = 0$.First,we find the derivative: $f'(x) = \cos x - \sin x$.Setting $f'(c) = 0$,we get $\cos c - \sin c = 0$.This implies $\cos c = \sin c$,which simplifies to $	an c = 1$.In the interval $[0, 2\pi]$,the values of $c$ for which $	an c = 1$ are $c = \frac{\pi}{4}$ and $c = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$.Thus,the values are $\frac{\pi}{4}, \frac{5\pi}{4}$.

Values of $c$ as per Rolle's theorem for $f(x)=\sin x+\cos x+6$ on $[0, 2\pi]$ are

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