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: Differentiate the function.
Given \[ f(x)=\tan^{-1}(\sin x-\cos x) \] Using the derivative formula \[ \frac{d}{dx}(\tan^{-1}u)=\frac{u'}{1+u^2} \] where \[ u=\sin x-\cos x \] Differentiate \(u\): \[ u'=\cos x+\sin x \] Thus \[ f'(x)=\frac{\cos x+\sin x}{1+(\sin x-\cos x)^2} \] Step 2: Simplify the denominator.
\[ (\sin x-\cos x)^2 = \sin^2 x+\cos^2 x-2\sin x\cos x \] \[ =1-2\sin x\cos x \] Thus \[ 1+(\sin x-\cos x)^2 = 2-2\sin x\cos x \] Since this expression is always positive, the sign of \(f'(x)\) depends only on the numerator.
Step 3: Determine the sign of the derivative.
\[ f'(x) \propto \cos x+\sin x \] Thus \[ \cos x+\sin x=0 \] \[ \sin x=-\cos x \] \[ \tan x=-1 \] \[ x=\frac{3\pi}{4} \] within the interval \((0,\pi)\).
Step 4: Determine increasing and decreasing intervals.
For \[ 0<x<\frac{3\pi}{4} \] \[ \cos x+\sin x>0 \] Thus \[ f'(x)>0 \] so the function is increasing.
For \[ \frac{3\pi}{4}<x<\pi \] \[ \cos x+\sin x<0 \] Thus \[ f'(x)<0 \] so the function is decreasing.
Step 5: Final result.
Hence the function increases in the first interval and decreases in the second interval.
Final Answer:
Increasing in \[ \boxed{\left(0,\frac{3\pi}{4}\right)} \] Decreasing in \[ \boxed{\left(\frac{3\pi}{4},\pi\right)} \]