For what values of x is the function f(x) = t | vedclass

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(D) To find where the function $f(x) = 	an^{-1}(\sin x + \cos x)$ is strictly increasing,we calculate its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\

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$0 < x < \pi/4$

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(D) To find where the function $f(x) = an^{-1}(\sin x + \cos x)$ is strictly increasing,we calculate its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$For the function to be strictly increasing,we must have $f'(x) > 0$.Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,the condition $f'(x) > 0$ simplifies to:$\cos x - \sin x > 0$$\cos x > \sin x$Dividing by $\cos x$ (assuming $\cos x > 0$),we get $ an x Alternatively,using the identity $\cos x - \sin x = \sqrt{2} \cos(x + \pi/4)$:$\sqrt{2} \cos(x + \pi/4) > 0$This holds when $-\pi/2 Subtracting $\pi/4$ from all sides:$-\pi/2 - \pi/4 $-3\pi/4 Comparing this with the given options,the interval $0 < x < \pi/4$ is a subset of the interval where the function is strictly increasing.

For what values of $x$ is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ strictly increasing?

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