How do you find the intervals of increasing and decreasing using the first derivative given #y=1/(x+1)^2#?

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Answer 1 · 2017-01-11T06:57:32

#y# is increasing for #(-oo, -1)# and decreasing for #(-1, +oo)#

#y = 1/(x+1)^2#

#= (x+1)^-2#

#y' = -2(x+1)^-3 * 1 #

#y'# represents the slope of #y# at an point #x# in its domain

Hence: #y# will be increasing when #y' >0#
and decreasing when #y'<0#

#:. y # will be increasing for #-2(x+1)^-3 >0#
#-> (x+1)<0# i.e. #x< -1#

and #y # will be decreasing for #-2(x+1)^-3 <0#
#-> (x+1)>0# i.e. #x> -1#

Hence: #y# is increasing for #(-oo, -1)# and decreasing for #(-1, +oo)#

This can be seen from the graph of #y# below:

graph{1/(x+1)^2 [-3.847, 1.626, -0.537, 2.202]}