Question

Show that \( f(x) = \frac{4 \sin x}{2 + \cos x} - x \) is an increasing function of \( x \) in \( \left[ 0, \frac{\pi}{2} \right] \).

Hint:

To check if a function is increasing, compute \( f'(x) \) and verify whether it is non-negative in the given interval.

Answer 1

Step 1: Compute the first derivative \( f'(x) \). The given function is: \[ f(x) = \frac{4 \sin x}{2 + \cos x} - x \] Differentiating term by term: \[ f'(x) = \frac{(2 + \cos x)(4 \cos x) - (4 \sin x)(-\sin x)}{(2 + \cos x)^2} - 1 \] 

Step 2: Simplify the derivative. \[ f'(x) = \frac{8 \cos x + 4 \cos^2 x + 4 \sin^2 x}{(2 + \cos x)^2} - 1 \] Using \( \sin^2 x + \cos^2 x = 1 \), we rewrite: \[ f'(x) = \frac{8 \cos x + 4}{(2 + \cos x)^2} - 1 \] \[ = \frac{8 \cos x + 4 - (2 + \cos x)^2}{(2 + \cos x)^2} \] Expanding \( (2 + \cos x)^2 = 4 + 4 \cos x + \cos^2 x \): \[ f'(x) = \frac{8 \cos x + 4 - (4 + 4 \cos x + \cos^2 x)}{(2 + \cos x)^2} \] \[ = \frac{8 \cos x + 4 - 4 - 4 \cos x - \cos^2 x}{(2 + \cos x)^2} \] \[ = \frac{4 \cos x - \cos^2 x}{(2 + \cos x)^2} \] 

Step 3: Check positivity in \( \left[ 0, \frac{\pi}{2} \right] \). For \( x \in \left[ 0, \frac{\pi}{2} \right] \), we know that \( \cos x \geq 0 \), so: \[ f'(x) = \frac{\cos x (4 - \cos x)}{(2 + \cos x)^2} \geq 0 \] since both \( \cos x \) and \( (4 - \cos x) \) are non-negative. 

Conclusion: Since \( f'(x) \geq 0 \) for all \( x \) in \( \left[ 0, \frac{\pi}{2} \right] \), \( f(x) \) is an increasing function in this interval.