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Thread: y=1-2x-x^2 Range?

  1. #1

    y=1-2x-x^2 Range?

    Please find the range of this using inverse function. Because inverse domain is function's range.

  2. #2
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    Are you trying to find the inverse domain by graphing, knowing that the inverse parabola is a reflection of the given parabola across the line y=x? You could reflect the coordinates of the given parabola's vertex, and reason with that.

    If you're trying to find the inverse algebraically, how far did you get? Please show your work. There are two functions for the inverse graph (one for each half of the reflected parabola), and they both share the same domain.


    PS: This comment is not related to your exercise. I just graphed the parabola and the inverse graphs. If the given function's domain is restricted from about -3.3 to 0.3, then its graph and the inverse graph form a Valentine.
    Last edited by Otis; 02-13-2019 at 06:04 PM. Reason: postscript and rewordings

  3. #3
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    You can't find the range "using inverse function" because [tex]f(x)=1- 2x- x^2[/tex] doesn't have an inverse function!

    You can, however, "complete the square": [tex]1- (x^2+ 2x+ 1- 1)= 1- (x^2+ 2x+ 1)+ 1= 1- (x+ 1)^2[/tex]. Since the square of a real number is always positive, that is 1 minus a positive number. That is, it is a parabola, opening downward from vertex (-1, 1). Its range is from negative infinity to 1.

  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    You can't find the range "using inverse function" because [tex]f(x)=1- 2x- x^2[/tex] doesn't have an inverse function!
    I had assumed sloppy wording, in the op (i.e., piecewise functions required).


    complete the square [tex]1 - (x^2 + 2x + 1) + 1 = 1 - (x+ 1)^2[/tex] [tex]+ \; 1[/tex]

    range is from negative infinity to 2.
    Two corrections shown in red.

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