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

5227 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-29T08:28:06

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{sinx-xcosx [-5.54, 14.46, -6.36, 3.64]}