Question

Find the sub–interval of \((0,\pi)\) in which the function \[ f(x)=\tan^{-1}(\sin x-\cos x) \] is increasing and decreasing.

Hint:

To find where a function increases or decreases, compute its derivative. If \(f'(x)>0\) the function is increasing, and if \(f'(x)<0\) the function is decreasing.

Answer 1

Step 1: Find the derivative of \( f(x) \). 
To determine where the function is increasing or decreasing, we need to find the first derivative of \( f(x) \). Since \( f(x) = \tan^{-1}(g(x)) \), we use the chain rule to differentiate: \[ f'(x) = \frac{1}{1 + [g(x)]^2} \cdot g'(x) \] where \( g(x) = \sin x - \cos x \). First, we find \( g'(x) \): \[ g(x) = \sin x - \cos x \quad \Rightarrow \quad g'(x) = \cos x + \sin x \] Therefore, the derivative of \( f(x) \) is: \[ f'(x) = \frac{1}{1 + (\sin x - \cos x)^2} \cdot (\cos x + \sin x) \] 

Step 2: Analyze the sign of \( f'(x) \). 
To determine where \( f(x) \) is increasing or decreasing, we need to analyze the sign of \( f'(x) \). The function \( f(x) \) will be increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \). The term \( 1 + (\sin x - \cos x)^2 \) in the denominator is always positive since it is of the form \( 1 + \text{something squared} \), which is always positive. Therefore, the sign of \( f'(x) \) depends on the numerator \( (\cos x + \sin x) \). We need to determine where \( \cos x + \sin x \) is positive and negative. 

Step 3: Find where \( \cos x + \sin x = 0 \). \br We solve for \( x \) where \( \cos x + \sin x = 0 \): \[ \cos x = -\sin x \] Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)): \[ 1 = -\tan x \quad \Rightarrow \quad \tan x = -1 \] The solution to \( \tan x = -1 \) in the interval \( (0, \pi) \) occurs at: \[ x = \frac{3\pi}{4} \] 

Step 4: Determine the intervals where \( f(x) \) is increasing and decreasing. 
In the interval \( (0, \frac{3\pi}{4}) \), \( \cos x + \sin x \) is positive because both \( \cos x \) and \( \sin x \) are positive in this interval. 2. In the interval \( (\frac{3\pi}{4}, \pi) \), \( \cos x + \sin x \) is negative because \( \cos x \) is negative and \( \sin x \) is also negative in this interval. Hence, we can conclude: - \( f(x) \) is increasing in the interval \( (0, \frac{3\pi}{4}) \) - \( f(x) \) is decreasing in the interval \( (\frac{3\pi}{4}, \pi) \) 

Final Answer: 
The function \( f(x) = \tan^{-1}(\sin x - \cos x) \) is: - Increasing in the interval \( (0, \frac{3\pi}{4}) \) - Decreasing in the interval \( (\frac{3\pi}{4}, \pi) \)