If #f(x)=sinx-xcosx#, how the function behaves in the intervals #(0,pi)# and #(pi,2pi)# i.e. whether it is increasing or decreasing?

Question

Aurora Alvarado · Accepted Answer

Between #(0,pi)#, #f(x)# is increasing and
between #(pi,2pi)#, #f(x)# is decreasing.

Whether a function #f(x)# is increasing or decreasing depends on whether #f'(x)=(df)/(dx)# is positive or negative. As #f(x)=sinx-xcosx# #f'(x)=(df)/(dx)=cosx-(1xxcosx+x xx(-sinx))# = #cosx-cosx+xsinx=xsinx# Now between #(pi,2pi)#, #sinx# is negative, but #x# is positive, hence #f'(x)# is negative, and between #(0,pi)#, #sinx# is positive and #x# too is positive, hence #f'(x)# is positive. Hence while between #(0,pi)#, #f(x)# is increasing, between #(pi,2pi)#, #f(x)# is decreasing. graph{-5.54, 14.46, -6.36, 3.64]}

Christopher Agresta · Answer

undefined To determine if $ f(x) = \sin(x) - x\cos(x) $ is increasing or decreasing in the intervals $(0,\pi)$ and $(\pi, 2\pi)$, we need to examine the derivative of $ f(x) $ in each interval.

1. Interval $(0,\pi)$:
   - Take the derivative of $ f(x) $: $ f'(x) = \cos(x) - \cos(x) + x\sin(x) = x\sin(x) $.
   - Since $ \sin(x) $ is positive in $(0,\pi)$ and $ x $ is also positive in this interval, $ f'(x) $ is positive.
   - Therefore, $ f(x) $ is increasing in $(0,\pi)$.

2. Interval $(\pi, 2\pi)$:
   - Again, take the derivative of $ f(x) $: $ f'(x) = x\sin(x) $.
   - In this interval, $ \sin(x) $ is negative, but $ x $ is positive.
   - Therefore, $ f'(x) $ is negative.
   - Consequently, $ f(x) $ is decreasing in $(\pi, 2\pi)$.