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Babynini

  • one year ago

f(x)= x/(1+x) g(x) = sin2x Find the following (along with their domains) a) fog b) gof c)fof d) gog

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  1. Babynini
    • one year ago
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    @jim_thompson5910 help please!

  2. Babynini
    • one year ago
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    a) fog f(g(x)) = f(sin2x)=sin2(1/(1+x))

  3. zepdrix
    • one year ago
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    Hmm that fog looks a little mixed up :d

  4. zepdrix
    • one year ago
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    \[\large\rm f(\color{orangered}{x})=\frac{\color{orangered}{x}}{1+\color{orangered}{x}}\]Becomes:\[\large\rm f(\color{orangered}{g(x)})=\frac{\color{orangered}{\sin2x}}{1+\color{orangered}{\sin2x}}\]Ya? :o

  5. zepdrix
    • one year ago
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    What you posted: \(\large\rm sin\left(2\color{orangered}{\frac{1}{1+x}}\right)\) This is actually the function f being plugged into g, \[\large\rm \sin\left(2\color{orangered}{\frac{1}{1+x}}\right)=\sin\left(2\color{orangered}{f(x)}\right)=g(\color{orangered}{f(x)})\]Where,\[\large\rm \sin(2\color{orangered}{x})=g(\color{orangered}{x})\]

  6. Babynini
    • one year ago
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    Just a second, trying to work it out on paper :)

  7. Babynini
    • one year ago
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    haha yeah I wrote it correctly on paper but did it wrong on here xD sorry! just a moment.

  8. zepdrix
    • one year ago
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    :3

  9. zepdrix
    • one year ago
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    For part A, do you understand how to find the domain of the composition?

  10. Babynini
    • one year ago
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    I am only having issues with the domain for c now. The rest are all entered and correct.

  11. Babynini
    • one year ago
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    c) x/1+2x Domain: (-infinity, -1/2) union (-1/2, infinity) it keeps saying the domain is wrong?

  12. zepdrix
    • one year ago
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    Oh sorry I ran off for a sec >.<

  13. Babynini
    • one year ago
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    You're all good :)

  14. zepdrix
    • one year ago
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    Hmm, I forget how these compositions work sometimes. I guess we need to carry this information around with us before simplifying,\[\large\rm f(x)=\frac{x}{1+x},\qquad\qquad x\ne-1\]So when we get to this composition we have,\[\large\rm f(f(x))=\frac{x}{1+2x},\qquad\qquad x\ne-1,-\frac{1}{2}\]Try that maybe? :o

  15. Babynini
    • one year ago
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    aaaah. so in interval notation it would be: (-infinity, -1) union (-1,-1/2)union(-1/2, infinity)

  16. zepdrix
    • one year ago
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    Mmmmm yah I think so :)

  17. zepdrix
    • one year ago
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    |dw:1443246434171:dw|

  18. Babynini
    • one year ago
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    Yaay! That is correct. Finally! Thank you so much :D I hadn't thought of that. So it is important to keep in mind the domains for the original functions

  19. zepdrix
    • one year ago
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    Yah, that's kinda tricky! :)

  20. Babynini
    • one year ago
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    You and your gorgeous drawings xP thanks. That makes sense.

  21. zepdrix
    • one year ago
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    XD

  22. zepdrix
    • one year ago
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    need moar light bulbs ...... to bury

  23. Babynini
    • one year ago
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    hahah we'll see.

  24. DanJS
    • one year ago
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    domain of nested functions like that is the intersection of each domain, for f(g(x)) , domain of g and domain of f of g.

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