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Math Help - inverse function help

  1. #1
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    inverse function help

    I dont know if this is right...

    i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

    so i got... x= -2(y+3)^2 +8
    x-8=-2(y+3)^2
    -1/2(x-8)= (y+3)^2
    square root both sides gives g-1(x)= √ -1/2(x-8)-3
    but this seems horribly wrong, - under a root and how do i find the domain from that??
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  2. #2
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    First you have to prove that g is bijective, if you do that the problem is like solved.

    "so i got... x= -2(y+3)^2 +8
    x-8=-2(y+3)^2
    -1/2(x-8)= (y+3)^2
    square root both sides gives g-1(x)= √ -1/2(x-8)-3"

    I don't understand - you should just calculate, then find \Delta (or do that when you prove that g is surjective) and the root that is in the interval you want it to be.
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  3. #3
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    Quote Originally Posted by dragon555 View Post
    I dont know if this is right...

    i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

    so i got... x= -2(y+3)^2 +8
    x-8=-2(y+3)^2
    -1/2(x-8)= (y+3)^2
    square root both sides gives g-1(x)= √ -1/2(x-8)-3
    but this seems horribly wrong, - under a root and how do i find the domain from that??
    g(x) = -2(x+3)^2 +8 , with domain -5 \le x \le -3

    range of g(x) is 0 \le y \le 8

    note that the range of  g(x) is the domain of  g^{-1}(x) ...

    x = -2(y+3)^2 +8

    x-8 = -2(y+3)^2

    \dfrac{8-x}{2} = (y+3)^2

    \pm \sqrt{\dfrac{8-x}{2}} = y+3

    y = \pm \sqrt{\dfrac{8-x}{2}} - 3

    for the domain 0 \le x \le 8 , you want the range to be -5 \le y \le -3

    g^{-1}(x) = -\sqrt{\dfrac{8-x}{2}} - 3
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  4. #4
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    Quote Originally Posted by dragon555 View Post
    I dont know if this is right...

    i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

    so i got... x= -2(y+3)^2 +8
    x-8=-2(y+3)^2
    -1/2(x-8)= (y+3)^2
    square root both sides gives g-1(x)= √ -1/2(x-8)-3
    but this seems horribly wrong, - under a root and how do i find the domain from that??
    It seesm that this question might be part of an assignment that counts towards your final grade. See rule #6: http://www.mathhelpforum.com/math-he...hp?do=vsarules. Thread closed.
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