Question

Show that the function \( f(x) = \sin x + \cos x \) is strictly decreasing in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \).

Hint:

To determine where a function is increasing or decreasing, compute the first derivative and analyze its sign. If the derivative is negative, the function is decreasing.

Answer 1

We are given the function \( f(x) = \sin x + \cos x \). 
To show that the function is strictly decreasing in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \), we first need to compute the derivative of \( f(x) \). 
Step 1: Find the derivative of \( f(x) \)
The derivative of \( f(x) = \sin x + \cos x \) is: \[ f'(x) = \cos x - \sin x \] 
Step 2: Solve for critical points by setting \( f'(x) = 0 \)
To find the critical points, solve the equation \( f'(x) = 0 \): \[ \cos x - \sin x = 0 \] \[ \cos x = \sin x \] 
This happens when: \[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4} \] 
Step 3: Determine the sign of \( f'(x) \) in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \)
Now, examine the sign of \( f'(x) = \cos x - \sin x \) in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \). 
- For \( x \in \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \), we know that \( \cos x < \sin x \), so \( f'(x) < 0 \). 
Since \( f'(x) < 0 \) in this interval, the function is strictly decreasing. 
Thus, we have shown that \( f(x) \) is strictly decreasing in the interval \( \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \).