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Domain, Range & Inverse of a Function

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    4ge9w.png
    How to solve part (iv) & (v)


    2. Relevant equations
    general form : [itex]y = a(x-h)^2 + k[/itex]


    3. The attempt at a solution
    In part (iv) for finding domain and range I converted g(x) in general form and then compared it with general form.
     
  2. jcsd
  3. Nov 7, 2012 #2

    symbolipoint

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    The inverse relation of y=8x-x2 is x=8y-y2.

    Also, the domain of the original relation is the range of the inverse relation in general. Specifics depend...
     
  4. Nov 7, 2012 #3

    HallsofIvy

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    I don't believe that is quite what is being asked. Completing the square, 8x- x2= 16- 16+ 8x- x2= 16- (x-4)2. The graph of that is a parabola having vertex at (4, 16). The function f(x)= 16- (x- 4)2 with [itex]x\le 4[/itex] has inverse function [itex]f^-1(x)= 4- \sqrt{x- 16}[/itex] while the function f(x)= 16- (x- 4)2 with [itex]x\ge 4 has inverse function [itex]f^-1(x)= 4+ \sqrt{x- 16}[/itex].

    By separating at the vertex, we cut the given function into to one-to-one that now have inverses.
     
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