The function f(x) = tan^{-1}(sin x + cos x),x | vedclass

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(C) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.We can rewrite the expression inside the inverse tangent as:$f(x) = 	an^{-1}\left(\sqrt{2} \left(\frac{

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$(0, \pi/4)$

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(C) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.We can rewrite the expression inside the inverse tangent as:$f(x) = 	an^{-1}\left(\sqrt{2} \left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos xight)ight) = 	an^{-1}\left(\sqrt{2} \sin\left(x + \frac{\pi}{4}ight)ight)$.To find where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{1}{1 + (\sqrt{2} \sin(x + \pi/4))^2} \cdot \sqrt{2} \cos(x + \pi/4)$.For $f(x)$ to be an increasing function,we require $f'(x) > 0$.Since the denominator $1 + 2 \sin^2(x + \pi/4)$ is always positive,we only need the numerator to be positive:$\sqrt{2} \cos(x + \pi/4) > 0$.This implies $\cos(x + \pi/4) > 0$.For $x > 0$,the cosine function is positive when the angle is in the first quadrant:$0 Subtracting $\frac{\pi}{4}$ from all sides,we get:$-\frac{\pi}{4} Given the condition $x > 0$,the interval is $(0, \pi/4)$.

The function $f(x) = \tan^{-1}(\sin x + \cos x)$,$x > 0$ is always an increasing function on the interval

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