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## Fanduekisses one year ago *How to find the domain of a composite function?

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1. Fanduekisses

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

2. Fanduekisses

Do I look at the radical or the denominator first?

3. anonymous

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

4. Fanduekisses

It can't equal 0

5. anonymous

right it has to be greater than 0

6. Fanduekisses

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

7. Fanduekisses

well, ok it doesn't matter lol

8. anonymous

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

9. Fanduekisses

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

10. Fanduekisses

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

11. anonymous

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)).|dw:1443058002457:dw|

12. anonymous

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

13. Fanduekisses

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...

14. Fanduekisses

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

15. anonymous

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|dw:1443058814223:dw|

16. anonymous

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

17. Fanduekisses

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

18. anonymous

Yes

19. Fanduekisses

Ohhhhhhhhhh

20. Fanduekisses

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

21. anonymous

exactly

22. Fanduekisses

Thanks!

23. anonymous

you're welcome

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