How do you find the domain & range for # f(x)= 3/ (sin3x + cos 3x)^2 -1#?

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Answer 1 · 2018-07-28T03:29:17

Range: # y notin ( 0, 0.5 )#.
Domain: #x ne ( 4 k - 1 ) pi/12, k = 0, +-1, +-2, +-3, ...#

#y = 3/( sin 3x + cos 3x )^2 -1 #

#= (3/2)/(1/sqrt2sin 3x +1/sqrt2 cos 3x )^2 - 1#

#= 1.5/(sin 3x cos (pi/4) + cos 3x sin (pi/4) )^2 - 1#

#= 1.5/sin^2( 3x +pi/4 ) -1 notin { 0, 0.5 )#, using

#csc^2# values #notin ( 0, 1 ),#

Asymptotes: {x = a zero of the denominator }

# rArr { 3 x + pi/4 = kpi}, k = 0, +-1, +-2, +-3, ..#

#rArr { x = ( 4 k - 1 ) pi/12 }#

# rArr x = ... -5/12pi, -pi/12, pi/4, ...

See graph.
graph{((y+1)(1+sin (6x))-3)(y-0.5 +0y)(x+1/12 pi+0.001y)(x-1/4 pi)=0}