Practise Now 1 The function \( f \) is defined by \( f(x)=\sin x+\cos x \) for \( 0 \leq x \leq 2 \pi \). Find the range of values of \( x \) for which \( f(x) \) is decreasing.

To determine where the function \( f(x) = \sin x + \cos x \) is decreasing, we first need to find its derivative, \( f'(x) \). The derivative of \( f(x) \) is: \[ f'(x) = \cos x - \sin x \] Next, we will find the values of \( x \) where \( f'(x) = 0 \): \[ \cos x - \sin x = 0 \] This simplifies to: \[ \cos x = \sin x \] This equation holds true when: \[ \tan x = 1 \] The solutions to this equation in the interval \( [0, 2\pi] \) are: \[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \] Thus, the solutions in our interval are: \[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4} \] To determine where \( f(x) \) is decreasing, we need to analyze the sign of \( f'(x) \): - For \( x < \frac{\pi}{4} \), \( f'(x) > 0 \) (increasing). - For \( \frac{\pi}{4} < x < \frac{5\pi}{4} \), \( f'(x) < 0 \) (decreasing). - For \( x > \frac{5\pi}{4} \), \( f'(x) > 0 \) (increasing). Thus, \( f(x) \) is decreasing in the interval: \[ \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \]