For what values of x is #f(x)= x-cosx# concave or convex?

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Answer 1 · 2016-01-16T17:02:09

#f(x)# is convex on #((-pi)/2+2kpi,pi/2+2kpi)# and concave on #(pi/2+2kpi,(3pi)/2+2kpi)# where #k# is an integer.

Concavity is determined by the sign of the second derivative:

First, determine the second derivative.

#f(x)=x-cosx#
#f'(x)=1+sinx#
#f''(x)=cosx#

So, we need to determine when #cosx# is positive and when it is negative. The sign of #cosx# will change whenever #cosx=0#.

This occurs when #x=(-pi)/2,pi/2,(3pi)/2,(5pi)/2#, and so on, increasing in intervals of #pi#.

#cosx>0# on #((-pi)/2+2kpi,pi/2+2kpi)# where #k# is an integer.
#cosx<0# on #(pi/2+2kpi,(3pi)/2+2kpi)# where #k# is an integer.

Thus,

#f(x)# is convex on #((-pi)/2+2kpi,pi/2+2kpi)# and concave on #(pi/2+2kpi,(3pi)/2+2kpi)# where #k# is an integer.