## JoannaBlackwelder one year ago Let f(x)=1/(x^2+1) and g(x)=x^-6. Find f(g(x)) and its domain.

1. JoannaBlackwelder

I got 1/((x^-6)^2 + 1) which simplifies to x^3/(1+x^3)

2. JoannaBlackwelder

That makes the domain all reals except for -1 the way I am looking at it, but that domain is not an option given.

3. JoannaBlackwelder

@kropot72

4. JoannaBlackwelder

The options for the domain are all real numbers x>0 x is not equal to 0 x is greater than or equal to 0

5. JoannaBlackwelder

Is it all real numbers since the powers were originally even?

6. JoannaBlackwelder

I thought the domain had to be real at each step though.

7. freckles

$f(x)=\frac{1}{x^2+1} \text{ and we have } g(x)=x^{-6} \text{ aka } g(x)=\frac{1}{x^6}$

8. JoannaBlackwelder

Right.

9. freckles

so notice the domains of f and g separately first

10. freckles

the domain of f is all real numbers but what is the domain of g? -- and also I'm having trouble seeing how you got your f(g(x)) anyways

11. JoannaBlackwelder

Domain of g is anything but 0

12. freckles

right @JoannaBlackwelder

13. JoannaBlackwelder

|dw:1441048089386:dw|

14. freckles

so we already know we can't include 0 in the domain of f(g(x))

15. freckles

right so you should have |dw:1441134566849:dw|

16. freckles

and you can get rid of the negative exponents by multiplying top and bottom by x^(12)

17. JoannaBlackwelder

Oh, duh! I don't know what I was thinking. Thank you!

18. kropot72

My simplification is: $\large \frac{1}{(x^{-6})^{2}+1}=\frac{x^{12}}{1+x^{12}}$

19. freckles

you do understand the domain is all real numbers but x=0 right?

20. JoannaBlackwelder

Yeah, I got it now. Thanks so much, both of you! :-)

21. freckles

this is because g(x) did not exist at x=0 you originally have $f(g(x))=\frac{1}{(\frac{1}{x^6})^{2}+1} \\ \text{ and pluggin \in 0 into this makes that } \\ \text{ one little fraction undefined }$