The sum of the absolute maximum and absolute | vedclass

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(C) Given $f(x) = 	an^{-1}(\sin x - \cos x)$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{1}{1 + (\sin x -

Vedclass · Accepted Answer

$	an^{-1}(\sqrt{2}) - \frac{\pi}{4}$

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(C) Given $f(x) = 	an^{-1}(\sin x - \cos x)$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{1}{1 + (\sin x - \cos x)^2} \cdot (\cos x + \sin x)$.Setting $f'(x) = 0$,we get $\cos x + \sin x = 0$,which implies $	an x = -1$.In the interval $[0, \pi]$,the only solution is $x = \frac{3\pi}{4}$.Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, \pi]$:$1$. At $x = 0$: $f(0) = 	an^{-1}(\sin 0 - \cos 0) = 	an^{-1}(0 - 1) = 	an^{-1}(-1) = -\frac{\pi}{4}$.$2$. At $x = \frac{3\pi}{4}$: $f\left(\frac{3\pi}{4}ight) = 	an^{-1}\left(\sin\frac{3\pi}{4} - \cos\frac{3\pi}{4}ight) = 	an^{-1}\left(\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})ight) = 	an^{-1}(\sqrt{2})$.$3$. At $x = \pi$: $f(\pi) = 	an^{-1}(\sin \pi - \cos \pi) = 	an^{-1}(0 - (-1)) = 	an^{-1}(1) = \frac{\pi}{4}$.Comparing these values:Absolute maximum value = $	an^{-1}(\sqrt{2})$.Absolute minimum value = $-\frac{\pi}{4}$.The sum of the absolute maximum and absolute minimum values is $	an^{-1}(\sqrt{2}) - \frac{\pi}{4}$.

The sum of the absolute maximum and absolute minimum values of the function $f(x) = \tan^{-1}(\sin x - \cos x)$ in the interval $[0, \pi]$ is.

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