In which interval is the function f(x) = tan^ | vedclass

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(B) To find the interval where the function $f(x) = 	an^{-1}(\sin x + \cos x)$ is increasing,we need to find the derivative $f'(x)$ and set it to be

Vedclass · Accepted Answer

$(-\pi/2, \pi/4)$

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(B) To find the interval where the function $f(x) = 	an^{-1}(\sin x + \cos x)$ is increasing,we need to find the derivative $f'(x)$ and set it to be greater than $0$.Given $f(x) = 	an^{-1}(\sin x + \cos x)$.Differentiating with respect to $x$ using the chain rule:$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} 	imes (\cos x - \sin x)$.For the function to be increasing,$f'(x) > 0$.Since $1 + (\sin x + \cos x)^2$ is always positive,we require:$\cos x - \sin x > 0$$\cos x > \sin x$Dividing by $\cos x$ (assuming $\cos x > 0$ in the relevant domain),we get:$1 > 	an x$$	an x Since $	an x = 1$ at $x = \pi/4$,the condition $	an x Considering the standard domain of the trigonometric functions involved,the interval is $(-\pi/2, \pi/4)$.

In which interval is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ an increasing function?

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