Question

Show that \( f(x) = \tan^{-1}(\sin x + \cos x) \) is an increasing function in \( \left[ 0, \frac{\pi}{4} \right] \).

Hint:

To check if a function is increasing, find its derivative and check if it is positive over the desired interval.

Answer 1

To show that the function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing, we need to show that the derivative \( f'(x) \) is positive in the interval \( \left[ 0, \frac{\pi}{4} \right] \). First, we differentiate \( f(x) \) with respect to \( x \). Using the chain rule: \[ f'(x) = \frac{d}{dx} \left( \tan^{-1}(\sin x + \cos x) \right) \] The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \), so: \[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx} (\sin x + \cos x) \] Next, we differentiate \( \sin x + \cos x \): \[ \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x \] Thus, the derivative of \( f(x) \) is: \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \] Now, we need to check the sign of \( f'(x) \) in the interval \( \left[ 0, \frac{\pi}{4} \right] \). In this interval, \( \cos x - \sin x \) is positive, since \( \cos x \) is greater than \( \sin x \) for \( x \in \left[ 0, \frac{\pi}{4} \right] \). Also, the denominator \( 1 + (\sin x + \cos x)^2 \) is always positive. Therefore, \( f'(x)>0 \) in \( \left[ 0, \frac{\pi}{4} \right] \), which implies that \( f(x) \) is an increasing function on this interval.