How do you find the domain of #g(x) = (3^(x+1)) / (sin x + cos x) #?

1 Answer
Feb 17, 2017

#{x| x!= (3pi)/4 + pin, x in RR, "where " n in ZZ}#

Explanation:

For this problem, we have to ask ourselves a couple of questions.

•When does #sinx + cosx# equal #0#.
•What is the domain of #3^(x + 1)#

I'll start by answering the first one. Solve the trigonometric equation.

#sinx + cosx = 0#

Square both sides

#(sinx + cosx)^2 = 0^2#

#sin^2x + 2sinxcosx + cos^2x = 0#

Apply #sin^2x + cos^2x = 1#:

#1 + 2sinxcosx = 0#

Use #2sinxcosx = sin2x#:

#1 + sin2x = 0#

#sin2x = -1#

#2x = arcsin(-1)#

#2x = (3pi)/2 + 2pin# because the sine function has a period of #2pi#

#x = (3pi)/4 + pin#

This means that whenever #x = (3pi)/4 + pin#, #n# an integer, the graph of #g(x) = 3^(x + 1)/(sinx + cosx)# will have vertical asymptotes.

Let's answer the second question.

#3^(x + 1)# is your run of the mill exponential function ; it will have a domain of all the real numbers, but will have a restricted range (we aren't dealing with range in this problem, though, so I won't go into detail there).

This means that the domain of #g(x)# is #{x| x!= (3pi)/4 + pin, x in RR, "where " n in ZZ}#.

Hopefully this helps!