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owlet

  • one year ago

help, I'm stuck with the question below. About Inverse Function. I just need help getting the inverse of the function.

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  1. owlet
    • one year ago
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  2. anonymous
    • one year ago
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    this has to be done algebraically?

  3. owlet
    • one year ago
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    yeah, idk how to do it

  4. owlet
    • one year ago
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    i'm not sure how to do the sqrt part along with the y^3 part

  5. anonymous
    • one year ago
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    not really sure how to go about solving the equation for y, but they're basically asking you to find where y = 11. I graphed it and backed out a solution that way. not exactly efficient

  6. owlet
    • one year ago
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    i tried doing that also. First I planned to solve the value of x when f(x)=11 , so that I could just switch it. But I'm stuck. We're not allowed to use any devices during exams/tests/quizzes, so I wanted to learn how to do this.

  7. anonymous
    • one year ago
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    I was thing a change of variable, but the equations I ended up with weren't much better than this one

  8. anonymous
    • one year ago
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    i have a suggestion

  9. anonymous
    • one year ago
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    set \[x^3+\sqrt{x+7}=11\] and solve for \(x\)

  10. anonymous
    • one year ago
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    you can do it using algebra, but i would not there is a simpler way

  11. owlet
    • one year ago
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    as i've said before, that's the first thing I tried. sqrt(x+7)=11-x^3 then if I square both sides,I will still get a not so good set up which confused me.

  12. anonymous
    • one year ago
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    since 11 is an integer, \(x+7\) must be a perfect square

  13. anonymous
    • one year ago
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    think of a number (gotta be pretty small) that when added to 7 gives a perfect square you will probably get it on the first try

  14. owlet
    • one year ago
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    7+2=9, so 2?

  15. anonymous
    • one year ago
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    an easy check now right?

  16. anonymous
    • one year ago
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    \[2^3+\sqrt{2+7}=?\]

  17. owlet
    • one year ago
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    8+3=11.. so its really 2 wait that's it??

  18. anonymous
    • one year ago
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    lol yeah, that's it

  19. owlet
    • one year ago
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    oh okay thanks.. it's like a trial and error method then

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spraguer (Moderator)
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