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anonymous

  • one year ago

For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

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  1. anonymous
    • one year ago
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    @Hero

  2. anonymous
    • one year ago
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    @Astrophysics

  3. anonymous
    • one year ago
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    @UsukiDoll

  4. campbell_st
    • one year ago
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    just start by swapping x and y so the original equation is \[y = (8 - 2x)^2\] do the swap \[x = (8 - 2y)^2\] now may y the subject of the equation

  5. anonymous
    • one year ago
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    So should i solve for x? @campbell_st

  6. campbell_st
    • one year ago
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    no you should solve for y... 1st step is to take the square root of both sides of the equation...

  7. anonymous
    • one year ago
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    y=4− x√ 2,4+ x√ 2

  8. campbell_st
    • one year ago
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    well the problem is that the original equation is a function, the inverse will be a relationship... and your solution seems ok...

  9. campbell_st
    • one year ago
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    the way you test the inverse is a function is to use the vertical line test after graphing the curve

  10. anonymous
    • one year ago
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    @UsukiDoll

  11. anonymous
    • one year ago
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    It doesnt match any of my answers

  12. campbell_st
    • one year ago
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    so do you have \[f^{-1}(x) = \frac{8 - \sqrt{x}}{2}\]

  13. anonymous
    • one year ago
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    idk i cant see my answer A lol it's probably that one than

  14. anonymous
    • one year ago
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    All my other answers are a + not a - and I cant see answer A

  15. UsukiDoll
    • one year ago
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    take the one with the + ? Idk.. but @campbell_st is right. you switch the signs of x and y and solve for y

  16. anonymous
    • one year ago
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    Is the final answer correct? If so I'll just chose A because everything else is a + after the 8

  17. UsukiDoll
    • one year ago
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    ?!!!! ok looks like I have to square root both sides or I won't go further lol

  18. anonymous
    • one year ago
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    haha I just want to make sure the answer is correct

  19. UsukiDoll
    • one year ago
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    \[x = (8 - 2y)^2\] [square both sides so we have y by itself] \[\sqrt{x}=8-2y\] ok so I'm just gonna stop at this point. I don't want to get busted :P but anyway we need y by itself so I ____________ 8 from both sides

  20. anonymous
    • one year ago
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    subtracr

  21. UsukiDoll
    • one year ago
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    yes \[\sqrt{x}-8=-2y\] so now what do we need to do to get y by itself?

  22. anonymous
    • one year ago
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    Add 2?

  23. UsukiDoll
    • one year ago
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    not quite. We can't use addition... try again

  24. anonymous
    • one year ago
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    divide?

  25. UsukiDoll
    • one year ago
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    yes we divide -2 on both sides.

  26. UsukiDoll
    • one year ago
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    \[\frac{\sqrt{x}-8}{-2}=y\] we can simplify further

  27. UsukiDoll
    • one year ago
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    for starters we can split this equation up.... \[\frac{\sqrt{x}-8}{-2}=y \rightarrow \frac{-\sqrt{x}}{2}+\frac{-8}{-2}\]

  28. anonymous
    • one year ago
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    So his answer was right than

  29. UsukiDoll
    • one year ago
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    so what's -8/-2 ?

  30. anonymous
    • one year ago
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    -4

  31. UsukiDoll
    • one year ago
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    a negative divided by a negative turns positive

  32. anonymous
    • one year ago
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    4

  33. UsukiDoll
    • one year ago
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    yeah.. and about his answer..I think he dropped the negative sign

  34. UsukiDoll
    • one year ago
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    \[\frac{-\sqrt{x}}{2}+4\]

  35. anonymous
    • one year ago
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    ok this is my final question tonight lol

  36. anonymous
    • one year ago
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    For the function f(x) = x2 - 12, find (f o f-1)(4)

  37. anonymous
    • one year ago
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    do we put 4 in all the x's?

  38. UsukiDoll
    • one year ago
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    hmmmm looks like we need to find f^{-1} first and plug that guy into f(x)

  39. UsukiDoll
    • one year ago
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    \[f(x) =x^2-12\] f(x) is also y \[y=x^2-12 \] now how do we find the inverse ?

  40. anonymous
    • one year ago
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    solve for y?

  41. UsukiDoll
    • one year ago
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    that's the second step what is the first step? what needs to be switched?

  42. anonymous
    • one year ago
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    the x and y

  43. UsukiDoll
    • one year ago
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    yes now we solve for y \[x=y^2-12\]

  44. UsukiDoll
    • one year ago
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    so how do we get y by itself? we need to __________ 12 to both sides

  45. anonymous
    • one year ago
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    add

  46. UsukiDoll
    • one year ago
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    \[x+12=y^2 \] good now we need to ___________ both sides we need a y by itself... it can't have an exponent

  47. anonymous
    • one year ago
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    divide?

  48. UsukiDoll
    • one year ago
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    no. umm.. what is the square root of y^2 ?

  49. anonymous
    • one year ago
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    nothing?

  50. UsukiDoll
    • one year ago
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    \[\sqrt{y^2} = ? \]

  51. UsukiDoll
    • one year ago
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    it can't be nothing.. otherwise we won't have y ._.

  52. UsukiDoll
    • one year ago
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    maybe exponential form will be easier \[\large y^\frac{2}{2}\] what's 2/2 ?

  53. anonymous
    • one year ago
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    1

  54. anonymous
    • one year ago
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    it would just turn into y

  55. UsukiDoll
    • one year ago
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    yes but keep in mind since we took the square root of y^2 we have to take the square root of the left hand side as well \[\sqrt{x+12}=y\]

  56. UsukiDoll
    • one year ago
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    eep \[f^{-1}(x) = \sqrt{x+12} \]

  57. UsukiDoll
    • one year ago
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    now we will place that inverse function into our \[f(x) = x^2-12 \] but something unusual is going to happen

  58. UsukiDoll
    • one year ago
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    oh wait it won't sorry we will still have something ok good whew

  59. anonymous
    • one year ago
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    haha ok

  60. UsukiDoll
    • one year ago
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    so f o f^{-1} means place f^{-1} inside the f(x) function.. yeah I was just looking at the number and not the variable xD

  61. anonymous
    • one year ago
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    Is the answer 6?

  62. UsukiDoll
    • one year ago
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    no try placing f inverse in f(x)

  63. anonymous
    • one year ago
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    4?

  64. UsukiDoll
    • one year ago
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    yeah

  65. anonymous
    • one year ago
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    sweetness

  66. anonymous
    • one year ago
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    im gonna need for you to be online whenever i am online lol

  67. UsukiDoll
    • one year ago
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    \[f^{-1}(x) = \sqrt{x+12} , f(x) = x^2-12 \] \[f(x) \cdot f^{-1}(x) \] \[f^{-1}(x) \cdot f(x) = (\sqrt{x+12})^2-12 \] \[f^{-1}(x) \cdot f(x) = x+12-12 \] \[f^{-1}(x) \cdot f(x) =x \] when x = 4 \[f^{-1}(x) \cdot f(x) =4 \]

  68. anonymous
    • one year ago
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    Thank you for all of your help, I got an A on my hw

  69. anonymous
    • one year ago
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    @UsukiDoll

  70. UsukiDoll
    • one year ago
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    oh that's awesome! :D

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