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emma97 Group Title

What is the inverse of f if….

  • 8 months ago
  • 8 months ago

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  1. emma97 Group Title
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    \[f(3) \sqrt[3]{x-2}\]

    • 8 months ago
  2. bibby Group Title
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    \[f(x)=\sqrt[3]{x-2}\] ?

    • 8 months ago
  3. emma97 Group Title
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    \[f(3)=\sqrt[3]{x-2}\]

    • 8 months ago
  4. emma97 Group Title
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    i forgot to put an = sign the first time

    • 8 months ago
  5. bibby Group Title
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    are we finding f'(3) either way to get the inverse, let's replace f(x) with y \[y = \sqrt[3]{x-2}\] switch x and y \[x = \sqrt[3]{y-2}\] cube both sides \[x^3 = {y-2}\] \[x^3+2=y\]

    • 8 months ago
  6. emma97 Group Title
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    \[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed

    • 8 months ago
  7. bibby Group Title
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    How'd you get y^3? Remember that square roots and squares are inverse operations, right? \[(\sqrt{x})^2 = x\] \[(\sqrt[3]{x})^3 = x\]

    • 8 months ago
  8. emma97 Group Title
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    oh i got confused i thought you said for ME to cube both sides lol. I'm sorry I'm extremely tired and have to get a bunch of work done by like tomorrow. ughhh. my bad

    • 8 months ago
  9. bibby Group Title
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    lol no worries, just always go over your work a bazillion times so that you can prove to yourself that your answer is right. ANYHOW, the inverse is now f(x) =x^2+2 f(3) = 3^2+2 = ??

    • 8 months ago
  10. emma97 Group Title
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    wait so the answer is f(x) =x^2+2 ?

    • 8 months ago
  11. bibby Group Title
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    whoops./ you generally say inverse with a little dash f'(x) so the equation for the inverse is f'(x) = x^2 + 2 then you plug in 3 into the inverse equation (assuming that's the question)

    • 8 months ago
  12. emma97 Group Title
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    heres the actual question with the answer choices so you can see

    • 8 months ago
  13. bibby Group Title
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    god damn. I kept saying x^2 +2 but I meant x^3 + 2

    • 8 months ago
  14. emma97 Group Title
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    hahah so is it the 3rd option then?

    • 8 months ago
  15. bibby Group Title
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    it is, it is...

    • 8 months ago
  16. emma97 Group Title
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    thank youuu! i have another question! do you mind helping me??

    • 8 months ago
  17. bibby Group Title
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    are you gonna post it here or make a new question? I don't mind

    • 8 months ago
  18. emma97 Group Title
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    ill post it here! and thanks! (:

    • 8 months ago
  19. emma97 Group Title
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    • 8 months ago
  20. bibby Group Title
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    uh you'd multiply by the conjugate. pardon my handwriting

    • 8 months ago
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  21. emma97 Group Title
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    i dont understand...

    • 8 months ago
  22. bibby Group Title
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    to simplify radicals you multiply by the conjugate pair which means flipping the sign

    • 8 months ago
  23. bibby Group Title
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    why do you have that answer selected?

    • 8 months ago
  24. emma97 Group Title
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    what do you mean that answer selected?

    • 8 months ago
  25. bibby Group Title
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    what are the answer choices?

    • 8 months ago
  26. emma97 Group Title
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    the ones with the bubble next to them on the link i posted!

    • 8 months ago
  27. bibby Group Title
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    rofl I didn't see it scrolled down a bit.

    • 8 months ago
  28. emma97 Group Title
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    hahahah its all good! (:

    • 8 months ago
  29. emma97 Group Title
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    so what is the answer?

    • 8 months ago
  30. myininaya Group Title
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    Perform the operation bibby suggested and then simplify.

    • 8 months ago
  31. bibby Group Title
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    you can use wolfram if you don't like learning http://www.wolframalpha.com/input/?i=%28sqrt%283%29-sqrt%286%29%29%2F%28sqrt%283%29%2Bsqrt%286%29%29

    • 8 months ago
  32. emma97 Group Title
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    its not that i don't like learning its that i am in a hurry and i have no idea whats going on.

    • 8 months ago
  33. myininaya Group Title
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    \[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] Do this multiplication here.

    • 8 months ago
  34. myininaya Group Title
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    Why are you in a hurry? Is this a test?

    • 8 months ago
  35. emma97 Group Title
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    idk how to..

    • 8 months ago
  36. emma97 Group Title
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    i had to go out of town so I'm behind on a bunch of work due like really soon

    • 8 months ago
  37. myininaya Group Title
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    So you don't know how to multiply something like this (a+b)(c+d)?

    • 8 months ago
  38. myininaya Group Title
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    A lot of Americans like to use this thing called the foil method.

    • 8 months ago
  39. emma97 Group Title
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    i have heard of it but idk how to really use it

    • 8 months ago
  40. myininaya Group Title
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    (a+b)(c+d)=ac+ad+bc+bd

    • 8 months ago
  41. emma97 Group Title
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    all you did was distribute those right?

    • 8 months ago
  42. myininaya Group Title
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    |dw:1388451041982:dw|

    • 8 months ago
  43. myininaya Group Title
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    yep|dw:1388451059969:dw|

    • 8 months ago
  44. myininaya Group Title
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    (a+b)(c+d)=a(c+d)+b(c+d) =ac+ad+bc+bd

    • 8 months ago
  45. myininaya Group Title
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    So if you had \[(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})\] what do you have after multiplying?

    • 8 months ago
  46. emma97 Group Title
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    idk how to group square roots together or multiply them together

    • 8 months ago
  47. myininaya Group Title
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    Do that foil method then we will work on simplifying

    • 8 months ago
  48. emma97 Group Title
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    hold on its gonna take awhile to plug in

    • 8 months ago
  49. myininaya Group Title
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    You don't have to use what I gave you as a formula. Just distribute. Take each of the terms in the first grouping and multiply to each term in the second grouping

    • 8 months ago
  50. emma97 Group Title
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    oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...

    • 8 months ago
  51. myininaya Group Title
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    Well I will show you how to do the first one and you try this one.

    • 8 months ago
  52. emma97 Group Title
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    i tried solving it and i think its the second answer choice

    • 8 months ago
  53. myininaya Group Title
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    \[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] This was the operation you were asked to perform. I was having you look at the numerator first. So looking at the numerator if we multiply the top out we would get: \[\frac{\sqrt{3} \sqrt{3}-\sqrt{3} \sqrt{6}-\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}\]

    • 8 months ago
  54. myininaya Group Title
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    Do you see how I got that top part?

    • 8 months ago
  55. emma97 Group Title
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    yes

    • 8 months ago
  56. myininaya Group Title
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    If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]

    • 8 months ago
  57. myininaya Group Title
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    so basically \[\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}\]

    • 8 months ago
  58. myininaya Group Title
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    combine like terms on top and multiply the bottom out i have to go for now

    • 8 months ago
  59. emma97 Group Title
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    ok.

    • 8 months ago
  60. Chooch146 Group Title
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    This has to be the longest problem eve

    • 8 months ago
  61. Chooch146 Group Title
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    ever* lol my name is eve:P

    • 8 months ago
  62. emma97 Group Title
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    hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh

    • 8 months ago
  63. Chooch146 Group Title
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    Hai!

    • 8 months ago
  64. emma97 Group Title
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    hahah whats up!

    • 8 months ago
  65. Chooch146 Group Title
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    not mch jst listening to music i was about to leave lol

    • 8 months ago
  66. emma97 Group Title
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    hahaha ok then! lol

    • 8 months ago
  67. Chooch146 Group Title
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    bye

    • 8 months ago
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