What is the relative maximum of y = csc (x)?

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Answer 1 · 2016-10-09T23:16:12

#y=cscx=1/sinx=(sinx)^-1#

To find a max/min we find the first derivative and find the values for which the derivative is zero.

#y=(sinx)^-1#
#:.y'=(-1)(sinx)^-2(cosx)# (chain rule)
#:.y'=-cosx/sin^2x#

At max/min, #y'=0=>-cosx/sin^2x=0#
#:.cosx=0#
#:.x=-pi/2,pi/2, ...#

When #x=pi/2=>y=1/sin(pi/2)=1#
When #x=-pi/2=>y=1/sin(-pi/2)=-1#

So there are turning points at #(-pi/2,-1)# and #(pi/2,1)#

If we look at the graph of #y=cscx# the we observe that #(-pi/2,-1)# is a relative maximum and #(pi/2,1)# is a relative minimum.

graph{csc x [-4, 4, -5, 5]}