Question

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

Answer 1

`{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!