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x3_drummerchick

  • one year ago

I dont know how to find the inverse of this! Someone help me get it started?? Will give medals: If g(x) = x^2 + 2x with x ≥ −1, find g^−1(15)=.

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  1. amilapsn
    • one year ago
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    find x in terms of g(x)

  2. x3_drummerchick
    • one year ago
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    i just dont know how to start it, we never learned hwoo to do inverse functions with restrictions

  3. amilapsn
    • one year ago
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    Here there's a restriction because, without the restriction \(g^{-1}(x)\) won't be a function...

  4. x3_drummerchick
    • one year ago
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    I know up to this point |dw:1442069050216:dw|

  5. amilapsn
    • one year ago
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    you're in the right path..

  6. x3_drummerchick
    • one year ago
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    it wouldnt be a one to one function, I understand that part. i just dont know how to fnd this inverse. do i factor x^2+2x or...?

  7. amilapsn
    • one year ago
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    do you know how to solve quadratic equations?

  8. x3_drummerchick
    • one year ago
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    yes

  9. x3_drummerchick
    • one year ago
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    |dw:1442069191193:dw|

  10. amilapsn
    • one year ago
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    not quite... Can you solve \(x^2+2x-y=0\) for x?

  11. x3_drummerchick
    • one year ago
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    im not sure how to do that one

  12. x3_drummerchick
    • one year ago
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    quadratic formula?

  13. amilapsn
    • one year ago
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    yep

  14. x3_drummerchick
    • one year ago
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    does i matter that there is a y in the equation? and do i go about doing it with a=1, b=2 and c=1?

  15. amilapsn
    • one year ago
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    you've to treat y as a constant..

  16. x3_drummerchick
    • one year ago
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    so a=1, b=2, c=-1; is tat why the restriction says that x>=1? cause the square root of it wouldve made the radical a negative; and you cant take the square root of negative number (its imaginary)

  17. x3_drummerchick
    • one year ago
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    |dw:1442069827280:dw|

  18. x3_drummerchick
    • one year ago
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    wait i messed that up: it should be

  19. x3_drummerchick
    • one year ago
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    |dw:1442069945844:dw|

  20. x3_drummerchick
    • one year ago
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    |dw:1442070088984:dw|

  21. x3_drummerchick
    • one year ago
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    |dw:1442070113092:dw|

  22. x3_drummerchick
    • one year ago
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    |dw:1442070133261:dw|

  23. x3_drummerchick
    • one year ago
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    so now what do i do, and is that correct? ^

  24. x3_drummerchick
    • one year ago
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    @Jhannybean help :(

  25. x3_drummerchick
    • one year ago
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    @amilapsn

  26. amilapsn
    • one year ago
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    c=-y

  27. x3_drummerchick
    • one year ago
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    i dont think this is right.. can you show some work? that doesnt make sense

  28. phi
    • one year ago
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    use the quadratic formula, with a=1, b=2, c= -y

  29. phi
    • one year ago
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    you should get \[ x= \frac{-2\pm \sqrt{4+4y}}{2} \] which simplifies to \[ x= -1 \pm \sqrt{1+y} \]

  30. phi
    • one year ago
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    now use the constraint that x \( \ge\) -1 that means we can't use the - sqr, otherwise x will be less than -1

  31. x3_drummerchick
    • one year ago
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    so that is the inverse of the function?

  32. x3_drummerchick
    • one year ago
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    oh okay t hat makes sense

  33. x3_drummerchick
    • one year ago
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    and so now I just plug in the 15 in for x?

  34. x3_drummerchick
    • one year ago
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    no wait, dont i have to solve for y first or something?

  35. phi
    • one year ago
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    so now you have a function (one value for every y) \[ x= -1 + \sqrt{1+y}\] we now "switch x and y" because we assume x is the "input" \[ y = -1 + \sqrt{1+x}\]

  36. x3_drummerchick
    • one year ago
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    ohhhhh okay

  37. phi
    • one year ago
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    now you have \[ g^{-1}(x) = -1 + \sqrt{1+x} \]

  38. x3_drummerchick
    • one year ago
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    thank you so much, that makes alot of sense

  39. x3_drummerchick
    • one year ago
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    my professor never showed us how to do these kinds -_- totally makes sense though, thanks again!

  40. phi
    • one year ago
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    the first trick was realizing you use the quadratic formula and treat -y as the "c"

  41. phi
    • one year ago
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    then the second idea is to use x>=-1 to know we don't use the -sqrt as the solution i.e. we use -1 + sqrt()

  42. x3_drummerchick
    • one year ago
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    yeah thats what threw me off because weve been doing pproblems where we just have to solve for 'y', but we didnt do a different approach to solving it, i didnt know we could do that with by putting a variable in the quad formula. that definitely threw me through a loophole lol

  43. x3_drummerchick
    • one year ago
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    and we cant use -1 because that would make a nonreal number- or imaginary number

  44. x3_drummerchick
    • one year ago
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    @phi quick question, how did you simplify under the radical??

  45. x3_drummerchick
    • one year ago
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    what did you do in these 2 steps?|dw:1442072831198:dw|

  46. x3_drummerchick
    • one year ago
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    |dw:1442072887637:dw|

  47. phi
    • one year ago
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    4+4y= 4(1+y) \[ \sqrt{4(1+y) }= \sqrt{4} \sqrt{1+y} \]

  48. x3_drummerchick
    • one year ago
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    oh wow, okay, thanks!

  49. phi
    • one year ago
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    btw \[ \sqrt{4+4y} \ne \sqrt{4}+ \sqrt{4y} \] for example, if y=1 we would have \[\sqrt{8} \ne 2+2 \\ \sqrt{8} \ne 4 \] as a calculator will show

  50. x3_drummerchick
    • one year ago
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    so those 2 are completely different, is it because its not held by multiplication?

  51. phi
    • one year ago
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    you can't split the sum inside a square root. you are allowed to factor into terms that multiply

  52. x3_drummerchick
    • one year ago
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    oh okay

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