Question

*How to find the domain of a composite function?

Answer 1

\[f(x)=\frac{ 1 }{ x^4-1 }\]
\[g(x)=\sqrt[6]{x}\]
\[g(f(x))=\sqrt[6]{\frac{ 1 }{ x^4-1 }}\]

Answer 2

Do I look at the radical or the denominator first?

Answer 3

I don't think it matters, but I guess start with the inside function. I think either way you're looking for where \(x^4-1>0\)

Answer 4

It can't equal 0

Answer 5

right it has to be greater than 0

Answer 6

because it's inside the radical or because it's in the denominator?

Answer 7

well, ok it doesn't matter lol

Answer 8

both. It can't be equal to 0 because of the denominator. It can't be less than 0 because of the even radical

Answer 9

Ok so, first the f(x), its domain is x cannot equal -1 or 1 ?

Answer 10

Then how do I find the domain of g(f(x)) ?

Answer 11

You put the 1 and -1 on a number line. Put numbers on either side of them into the function to see where its positive. So find g(f(-2)), g(f(0)), and g(f(2)).

Answer 12

really you're just looking for the sign since the inequality is asking for positive values. Wherever it's positive will be the domain

Answer 13

erhh I'm confused. :( or I'm exhausted, Idk why I'm finding this so confusing and it's probably not that hard to understand...

Answer 14

Can you please explain to me the rules of finding domain of a composition function?

Answer 15

There aren't any specific rules for composite functions. You have to look at the function once it's composed and try to figure out values excluded from the domain. For this function there's both a radical and rational parts. So it turns out that you'll have to solve a polynomial inequality to get the values.

Plugging in -2, 0, and 2 into \(x^4-1\).
\((-2)^4-1=15\)
\(0^4-1=-1\)
\(2^4-1=15\)
Put the sign on the number line

Answer 16

So since you're looking for the positive part, the domain is (-∞, -1) U (1, ∞)

Answer 17

So that is why you look at the domain of f(x) (inside) first? As like a guide?

Answer 18

Yes

Answer 19

Ohhhhhhhhhh

Answer 20

It makes so much sense now, like the domain has to work for both functions and stuff.

Answer 21

exactly

Answer 22

Thanks!

Answer 23

you're welcome