Question

g(x) = arcsin(4x + 8)
Find the domain of this function.

Answer 1

First, what is the domain of the function f(x) = arcsin(x) ?

Answer 2

I know that the domain of h(x) = arcsin is [-1,1]. So, if I were to change it to the domain of g(x), wouldn't I just take each of the end points of the domain and apply the following:

g(x) = arcsin(4(x+2))
So, to -1 and 1 individually, move left by 2 units, and multiply by 1/4, because they are horizontally stretched by a factor of 1/4?

Answer 3

So\[-1 \leq 4x + 8 \leq 1\]

Answer 4

The solution of that relation is the domain of your function

Answer 5

What? How did you get that?

Answer 6

The domain of arcsin(x) = [-1,1]

Answer 7

The domain of f(y) = arcsin(y) is y in [-1,1]
i.e.,
\[-1 \leq y \leq 1\]

Answer 8

I am confused as to what the y is doing ther.e

Answer 9

Here y = y(x) = 4x + 8. Hence
\[-1 \leq 4x + 8 \leq 1 \]

Answer 10

So only the terms within the argument of the function affect its domain?

Answer 11

y is the range
x is the domain

Answer 12

One way to think about this function g is that it is the composite of two functions: f(x) = arcsin(x) and h(x) = 4x + 8

You function g(x) = f(h(x))

Answer 13

So we need the RANGE of h(x) to be in the DOMAIN of f(x).

The domain of f(x) = arcsin(x) is as we've observed [-1,1]

Answer 14

Ok, so then how do you settle with the domain.

Answer 15

So we need the range of h(x) which is 4x + 8 for given x to satisfy that relation as well.

Answer 16

Hence solve
\[-1 \leq 4x + 8 \leq 1\]
to find the x for which the range of h is in the domain f. I.e., the x which is the domain for g(x) = arcsin(4x+8)

Answer 17

Exactly right.

Answer 18

So you are saying that where in the inner function does it work if it is to be composed? In this case, the inner functions' values are restricted from -1 to 1 because of the outer function's domain restrictions?

Answer 19

yes, more or less, yes.

Answer 20

For example, if x = 1, then 4x + 8 =12 and we can't evaluate
arcsin(4x+8)