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marcelie

  • one year ago

help please !!!! find the inverse f(x) 3x^2+5 please check my work

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  1. marcelie
    • one year ago
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    |dw:1443237553803:dw|

  2. steve816
    • one year ago
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    that is some really messed up handwriting xD

  3. marcelie
    • one year ago
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    the answer in my textbook says it -

  4. steve816
    • one year ago
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    I wish I can help you, but it is too hard to read my friend.

  5. marcelie
    • one year ago
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    sorry lol can you solve it ?

  6. steve816
    • one year ago
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    But to find the inverse, simply switch the x and y and solve for y.

  7. marcelie
    • one year ago
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    k one sec let me rewrite it

  8. steve816
    • one year ago
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    subtract 5, divide by 3, square root it.

  9. marcelie
    • one year ago
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    |dw:1443237871401:dw|

  10. marcelie
    • one year ago
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    |dw:1443238092545:dw|

  11. marcelie
    • one year ago
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    why does my textbook answer has the negative infront of the equation

  12. steve816
    • one year ago
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    Because you messed up a little bit.

  13. marcelie
    • one year ago
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    oh

  14. marcelie
    • one year ago
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    im confused ..

  15. steve816
    • one year ago
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    Don't be. Life has many confusing moments.

  16. marcelie
    • one year ago
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    yeah

  17. steve816
    • one year ago
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    But never give up and keep trying until you are successful!

  18. marcelie
    • one year ago
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    \[f (x ) = 3x ^{2}+5\]

  19. marcelie
    • one year ago
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    \[x= 3y ^{2}+5\]

  20. steve816
    • one year ago
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    3y^2 = x-4

  21. steve816
    • one year ago
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    \[y = \frac{ \sqrt{x-5} }{ \sqrt{3} }\]

  22. marcelie
    • one year ago
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    so do i first divide then square root it ? because i always get confused

  23. steve816
    • one year ago
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    You have to rationalize the denominator but I'm too lazy you can do that :)

  24. marcelie
    • one year ago
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    isnt the answer that ?

  25. steve816
    • one year ago
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    no, you can't have a square root in the denominator.

  26. marcelie
    • one year ago
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    mmk bc my solution has that answer and i dont understand why has negative

  27. marcelie
    • one year ago
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  28. marcelie
    • one year ago
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    so thats what i did

  29. DanJS
    • one year ago
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    \[\sqrt{\frac{ a }{ b }}=\frac{ \sqrt{a} }{ \sqrt{b} }\]

  30. DanJS
    • one year ago
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    recall, when you solve y = x^2 x = +root(y) or x= - root(y)

  31. marcelie
    • one year ago
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    oh , so then i have to write -/+

  32. marcelie
    • one year ago
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    soo every inverse has to have that -/+ when i take the sq root ?

  33. DanJS
    • one year ago
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    no, that is just a property of square roots

  34. DanJS
    • one year ago
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    but you have to figure which is the right answer

  35. marcelie
    • one year ago
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    oh bc most of the problems ive done are sq roots but my solotuin book doesnt include -/+

  36. marcelie
    • one year ago
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    how do i figure that out

  37. DanJS
    • one year ago
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    also another thing to note, \[\sqrt{y^2} = \left| y \right|\] the square root of y^2 will always be positive, and is really the absolute value of y, not just y

  38. DanJS
    • one year ago
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    and a 3rd recall, the absolute value function is defined |dw:1443239752302:dw|

  39. DanJS
    • one year ago
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    that is where the negative comes in

  40. DanJS
    • one year ago
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    Also as an aside, since you are (x,y) --->(y,x) swapping y and x, The inverse, if there is one, is a reflection over the line y = x ; through the origin slope 1

  41. DanJS
    • one year ago
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    notice the intervals they specified, they include all real numbers, but why break it up to that...

  42. DanJS
    • one year ago
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    Do they give just the negative root as an answer or both + and - roots depending on which interval you are talking about in the domain

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