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JoannaBlackwelder

  • one year ago

Let f(x)=1/(x^2+1) and g(x)=x^-6. Find f(g(x)) and its domain.

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  1. JoannaBlackwelder
    • one year ago
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    I got 1/((x^-6)^2 + 1) which simplifies to x^3/(1+x^3)

  2. JoannaBlackwelder
    • one year ago
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    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
    • one year ago
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    @kropot72

  4. JoannaBlackwelder
    • one year ago
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    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
    • one year ago
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    Is it all real numbers since the powers were originally even?

  6. JoannaBlackwelder
    • one year ago
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    I thought the domain had to be real at each step though.

  7. freckles
    • one year ago
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    \[f(x)=\frac{1}{x^2+1} \text{ and we have } g(x)=x^{-6} \text{ aka } g(x)=\frac{1}{x^6}\]

  8. JoannaBlackwelder
    • one year ago
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    Right.

  9. freckles
    • one year ago
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    so notice the domains of f and g separately first

  10. freckles
    • one year ago
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    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
    • one year ago
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    Domain of g is anything but 0

  12. freckles
    • one year ago
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    right @JoannaBlackwelder

  13. JoannaBlackwelder
    • one year ago
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    |dw:1441048089386:dw|

  14. freckles
    • one year ago
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    so we already know we can't include 0 in the domain of f(g(x))

  15. freckles
    • one year ago
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    right so you should have |dw:1441134566849:dw|

  16. freckles
    • one year ago
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    and you can get rid of the negative exponents by multiplying top and bottom by x^(12)

  17. JoannaBlackwelder
    • one year ago
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    Oh, duh! I don't know what I was thinking. Thank you!

  18. kropot72
    • one year ago
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    My simplification is: \[\large \frac{1}{(x^{-6})^{2}+1}=\frac{x^{12}}{1+x^{12}}\]

  19. freckles
    • one year ago
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    you do understand the domain is all real numbers but x=0 right?

  20. JoannaBlackwelder
    • one year ago
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    Yeah, I got it now. Thanks so much, both of you! :-)

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

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