## Fanduekisses one year ago *****WHY U NO HELP ME??!!

1. Fanduekisses

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

2. Fanduekisses

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

3. Fanduekisses

The radical and exponent part is confusing, how will I go about that?

4. Fanduekisses

@ganeshie8

5. Nnesha

you should convert radical to an exponent form

6. Nnesha

$\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }$index become the denomiantor of the fraction

7. Nnesha

denominator *

8. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Fanduekisses $f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1$ $$\color{blue}{\text{End of Quote}}$$ btw, -1 should be in the denominator

9. Nnesha

$\huge\rm \frac{ 1 }{(\color{ReD}{ \sqrt[6]{x}} )^4} =\frac{ 1 }{ (\color{Red}{x^{??}})^4 }$

10. Fanduekisses

$\frac{ 1 }{ (x ^{1/6} )^4}$

11. Nnesha

yes right now you can multiply 1/6 by 4 exponent rule $\huge\rm (a^m)^x = a^{m \times x}$

12. Fanduekisses

$\frac{ 1 }{ x ^{2/3} }$

13. Nnesha

don't forget the negative one

14. Fanduekisses

ok so the domain will be $x \ge0$ and

15. Fanduekisses

Wait how do I solve for x? $x ^{2/3}= 1$

16. Fanduekisses

The denominator part, to find the domain, what x cannot equal.

17. Nnesha

ugh sorry my laptop is acting up,slow net connection alright so whats the question ? find domain ?

18. Fanduekisses

yes of the composition f(g(x))

19. Nnesha

you can take cube both sides to get rid of the fraction

20. Nnesha

like we rewrite x^2/3 into radical form $\huge\rm \sqrt[3]{x^2}=1$ to cancel out cube root we should take cube sides just like we take square root to cancel out square

21. Fanduekisses

wait be right back

22. Nnesha

btw, you can apply difference of squres....

23. mathmate

$$\frac{ 1 }{ (\sqrt[6]{x})^4 }-1=\frac{1}{x^{2/3}}-1=x^{-2/3}-1=0$$ It doesn't change the answer in this case. lol

24. Fanduekisses

So th$x \neq1 ?$e domain of f(g(x)) is $x \ge0$ and

25. Fanduekisses

26. Nnesha

how many solutions would you get when u take square root of something ? and it's okay :=)

27. Fanduekisses

+- 1 hehe

28. Fanduekisses

:D got it now hehe thanks so much!

29. Fanduekisses

So the domain is x cannot equal +-1 and x has to be greater than or equal to 0 ?

30. Nnesha

to write in interval u can draw a number line |dw:1443037013961:dw|

31. Nnesha

|dw:1443037153659:dw|

32. Fanduekisses

$(-\infty, -1) U (-1, 1)(1,\infty)$

33. Nnesha

$(-\infty, -1) U (-1, 1) \rm U(1,\infty)$

34. Nnesha

there is another way to test each number if it's satisfy or not in the original equation i don't think that's necessary for that question hm

35. Fanduekisses

I forgot the union lol

36. Fanduekisses

thanks

37. Nnesha

ye

38. Nnesha

np here is good notes to practice on these stuff http://math.arizona.edu/~rwilliams/math112-spring2012/Notes/Domain_of_a_Composition.pdf np that was a good review for me actlly :/