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anonymous

  • one year ago

find inverse of y=(x+3)^2

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  1. anonymous
    • one year ago
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    @madhu.mukherjee.946

  2. Nnesha
    • one year ago
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    two equal signs ?

  3. Nnesha
    • one year ago
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    (x+3)^2 or (x-3)^2 ??

  4. anonymous
    • one year ago
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    (x+3)^2

  5. Nnesha
    • one year ago
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    okay cool same like last one (x+3)^2 is same as (x+3)(x+3) so apply the foil method to expand (x+3)^2 and then replace x with y and y with x

  6. madhu.mukherjee.946
    • one year ago
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    @Nnesha pls help me with a stats question i'm stuck i've already tagged u

  7. Nnesha
    • one year ago
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    stat >.>

  8. anonymous
    • one year ago
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    so i foiled (x+3)^2 I got x^2+6x+9 so if you switch the x and y you would get x=y^2+6y-9

  9. Nnesha
    • one year ago
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    wait a sec i said something wrong

  10. dinamix
    • one year ago
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    \[f(x)^{-1} =\sqrt{x}-3 \]for x>3 and \[f(x)^{-1} =-\sqrt{x}-3\] for x<3

  11. anonymous
    • one year ago
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    how do yo get that answer @dinamix

  12. dinamix
    • one year ago
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    @sarahm22

  13. dinamix
    • one year ago
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    only i use my mind cuz its easy for me this things

  14. dinamix
    • one year ago
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    @Nnesha did u see my answer

  15. Nnesha
    • one year ago
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    No.

  16. anonymous
    • one year ago
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    what are the steps to getting the answer

  17. dinamix
    • one year ago
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    for ops -3 not 3 sorry i type cery fast

  18. Nnesha
    • one year ago
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    how he got the answer: take square root both sides

  19. Nnesha
    • one year ago
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    that's what he did: replace x with y and y with x 2nd) then take square root both sides to cancel out the square

  20. dinamix
    • one year ago
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    \[\sqrt{[x]^2} = |x|\]

  21. dinamix
    • one year ago
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    did u know this rule right ?

  22. dinamix
    • one year ago
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    @Nnesha

  23. Nnesha
    • one year ago
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    yes ?

  24. dinamix
    • one year ago
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    \[\sqrt{(x)^2} = |x|\] did u know this rule @Nnesha

  25. Nnesha
    • one year ago
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    yes. I said NO .cuz i didn't know how you got +3

  26. dinamix
    • one year ago
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    \[y = (x+3)^2---> \sqrt{y}=|x+3| --> x=\sqrt{y}-3 --> f(x)^{-1} = \sqrt{x}-3\]

  27. dinamix
    • one year ago
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    @Nnesha

  28. dinamix
    • one year ago
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    this for only when x>-3

  29. dinamix
    • one year ago
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    did u see my steps ?

  30. Nnesha
    • one year ago
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    yes \[\huge\rm \sqrt{ x}=\pm \sqrt{(y+3)^2}\] just in case if he isn't familiar with the absolute value

  31. dinamix
    • one year ago
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    so my method is wrong @Nnesha o.O ?

  32. Nnesha
    • one year ago
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    i didn't say that :=)

  33. dinamix
    • one year ago
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    so what solution @Nnesha i think i am right cuz no mistakes

  34. Nnesha
    • one year ago
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    yes i think so. good job and thanks

  35. dinamix
    • one year ago
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    u know how do it for x<-3 right ?@Nnesha

  36. dinamix
    • one year ago
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    @Nnesha

  37. Nnesha
    • one year ago
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    yes i do

  38. dinamix
    • one year ago
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    @freckles my method is right ? look my answer plz

  39. dinamix
    • one year ago
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    its not 3 i forget - okey

  40. freckles
    • one year ago
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    I guess he was looking for inverse relation since an inverse function doesn't exist. And yeah... \[y=(x+3)^2 \\ \text{ take square root of both sides } \\ \pm \sqrt{y}=x+3 \\ \text{ now if } x+3 \text{ is positive or zero } (x+3 \ge 0) \\ \text{ then } x+3=\sqrt{y} \\ \text{ and if } x+3 \text{ is negative } \\ \text{ then } x+3=-\sqrt{y} \] \[\text{ solving each } \\ x=\sqrt{y}-3 \text{ if } x+3 \ge 0 \text{ or solving the inequality we get } x \ge -3 \\ x=-\sqrt{y}-3 \text{ if } x+3 <0 \text{ or solving that inequality we get } x<-3 \\ \text{ so we have } y=\sqrt{x}-3 \text{ if } x \ge -3 \\ \text{ or } y=-\sqrt{x}-3 \text{ if } x<-3 \] which you already said

  41. Nnesha
    • one year ago
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    i didn't say tht ur answer is incorrect ....

  42. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @dinamix its not 3 i forget - okey \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @dinamix for ops -3 not 3 sorry i type cery fast \(\color{blue}{\text{End of Quote}}\) i know.

  43. dinamix
    • one year ago
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    lol why do that @Nnesha

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