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anonymous

  • one year ago

if h(x)=2-x/2, find (h o h^-1)(2) A)X B)1/2 C)2 D)0 THANK YOU SO MUCH!!

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  1. jim_thompson5910
    • one year ago
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    is the function this? \[\Large h(x) = 2 - \frac{x}{2}\] OR is the function this? \[\Large h(x) = \frac{2-x}{2}\]

  2. anonymous
    • one year ago
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    No, it is..

  3. anonymous
    • one year ago
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    |dw:1442458455816:dw|

  4. jim_thompson5910
    • one year ago
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    ok so the first one I wrote

  5. anonymous
    • one year ago
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    So yes, the first one you stated. Thank you.

  6. jim_thompson5910
    • one year ago
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    think of it as \[\Large y = 2 - \frac{x}{2}\]

  7. jim_thompson5910
    • one year ago
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    then swap x and y \[\Large x = 2 - \frac{y}{2}\] then solve for y

  8. anonymous
    • one year ago
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    How

  9. anonymous
    • one year ago
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    what about the h o h ^-1(2)

  10. jim_thompson5910
    • one year ago
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    I would first multiply both sides by 2 that clears out the fraction

  11. jim_thompson5910
    • one year ago
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    so you'll have 2x = 4 - y isolate y

  12. anonymous
    • one year ago
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    -y=2x-4

  13. anonymous
    • one year ago
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    then what happens with the h o h ^1 (2). That is my primary concern.

  14. jim_thompson5910
    • one year ago
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    -y=2x-4 will turn into y = -2x+4

  15. jim_thompson5910
    • one year ago
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    so the inverse is \[\Large h^{-1}(x) = -2x+4\]

  16. jim_thompson5910
    • one year ago
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    \[\Large (h \circ h)^{-1}(2)\] is the same as \[\Large h(h^{-1}(2))\] the first task is to compute \(\Large h^{-1}(2)\)

  17. anonymous
    • one year ago
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    Okay

  18. jim_thompson5910
    • one year ago
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    sorry I meant to say \(\Large (h \circ h^{-1})(2)\)

  19. jim_thompson5910
    • one year ago
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    anyways, what output do you get when you plug x = 2 into the inverse

  20. anonymous
    • one year ago
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    0?

  21. jim_thompson5910
    • one year ago
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    yep

  22. anonymous
    • one year ago
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    Oh I see the -1

  23. jim_thompson5910
    • one year ago
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    then that 0 is plugged into the h(x) function

  24. jim_thompson5910
    • one year ago
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    \[\Large \Large h(\color{red}{h^{-1}(2)}) = h(\color{red}{0})\]

  25. anonymous
    • one year ago
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    Thank you so much for the walkthrough, it really helped me. :)

  26. jim_thompson5910
    • one year ago
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    so you found what h(0) is?

  27. anonymous
    • one year ago
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    0 right?

  28. jim_thompson5910
    • one year ago
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    plug it into 2 - (x/2)

  29. anonymous
    • one year ago
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    h^-1 (2)= -2(2)+4

  30. jim_thompson5910
    • one year ago
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    yeah that part is 0, but we're not done yet

  31. anonymous
    • one year ago
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    oh, so it is just 2?

  32. jim_thompson5910
    • one year ago
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    you take that result and plug it into h(x)

  33. jim_thompson5910
    • one year ago
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    ultimately, the final result is 2 you'll find that \[\Large (h \circ h^{-1})(x) = x\] for any x value

  34. anonymous
    • one year ago
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    2- 0/2=2-0

  35. jim_thompson5910
    • one year ago
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    because the inverse essentially "undoes" what the original function does

  36. anonymous
    • one year ago
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    Oh, I see.

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