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dtan5457

  • one year ago

if f(g(x)=2x/(x+z), and g(x)=2x/(x+4) find f(x)

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  1. dtan5457
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    is the `x+z` supposed to be `x+4` ?

  3. dtan5457
    • one year ago
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    they are both under 2x so no

  4. jim_thompson5910
    • one year ago
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    then I'm not sure. That z is throwing me off

  5. dtan5457
    • one year ago
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    perhaps its an x+2? I mean my teacher's handwriting does look like a Z.

  6. jim_thompson5910
    • one year ago
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    that sounds possible

  7. anonymous
    • one year ago
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    i have an answer that might work, but i think you are supposed to use inverse functions some how

  8. anonymous
    • one year ago
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    by trial and error i came up with \[f(x)=\frac{1}{x-\frac{2}{x}}\]] we can see if that works, and also clean it up a bit

  9. anonymous
    • one year ago
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    in any case it is wrong, but i think i can fix it

  10. anonymous
    • one year ago
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    lets try this one \[f(x)=\frac{1}{\frac{1}{x}-\frac{2}{2x}}\]

  11. anonymous
    • one year ago
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    ok one more time without me being so stupid \[f(x)=\frac{1}{\frac{1}{x}-\frac{2}{x+4}}\]

  12. Nnesha
    • one year ago
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    *

  13. anonymous
    • one year ago
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    i think that actually works

  14. dtan5457
    • one year ago
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    where would a Z come up though? or is that for x+2

  15. anonymous
    • one year ago
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    i used 2 with z it is too weird

  16. dtan5457
    • one year ago
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    so it would be impossible with z?

  17. anonymous
    • one year ago
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    i am wagering it is 2

  18. anonymous
    • one year ago
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    i wonder if we can do it with differentiation though, is that part of this class?

  19. dtan5457
    • one year ago
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    this is pre calc we only did difference quotients so far

  20. anonymous
    • one year ago
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    maybe @Zarkon has a better idea seems like it needs to use inverse functions some how but i am clogged up in any case i am almost certain the (third) answer i wrote above works, just trying to figure how to subtract 2 in the denominator

  21. anonymous
    • one year ago
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    take the reciprocal, subtract 2 in the numerator, then take the reciprocal again of course that mess i wrote can be cleaned up without using compound fractions

  22. dtan5457
    • one year ago
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    my teacher may have wrote something wrong of some sort its only been like 2 weeks into school i doubt he would have something harder than you can solve

  23. anonymous
    • one year ago
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    usually you are told \(f\) and \(f\circ g\) and asked to find \(g\) that you can do by finding the inverse of \(f\) and applying it this is a bit different

  24. dtan5457
    • one year ago
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    you are right in my class example we had to find g and that was fairly simple

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