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Math Help - inverse functions

  1. #1
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    inverse functions

    How do I find the inverse functions from abs(x^2 - 3)? WolframAlpha gives 4 expressions.
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  2. #2
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    Re: inverse functions

    This will not have an inverse function - it's not one-to-one... You could restrict the domain to get an inverse function though...
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  3. #3
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    Re: inverse functions

    Quote Originally Posted by Stuck Man View Post
    How do I find the inverse functions from abs(x^2 - 3)? WolframAlpha gives 4 expressions.
    Yes, it does! And the reason is exactly what Prove It said- the function is not "one to one". If you were to graph it you would see that there are some horizontal lines that cross the graph 4 times- so 4 different values of x that give the same y value.

    Specifically, y= x^2- 3 is a parabola that crosses the x-axis at (-\sqrt{3}, 0), goes down to (0, -3), back up to the x-axis at (\sqrt{3}, 0), and then up. Taking the absolute value then "folds" that portion below the x-axis up above it.

    If x\le -\sqrt{3}, then x^2\ge 3 so x^2- 3\ge 0. Taking the absolute value of a positive number doesn't change it so, for x\le -\sqrt{3}, the function is y= x^2- 3. To find the inverse function, swap x and y and solve for y: x= y^2- 3 so y^2= x+ 3 and y= \pm\sqrt{x+ 3}. But in this case, y (which was x but got swapped) is less than or equal to -3. y is negative so y= -\sqrt{x+ 3}.

    If -\sqrt{3}< x\le 0, then x^2< 3 so x^2- 3< 0. Taking the absolute value multiplies that by -1 so, for -\sqrt{3}< x< 0, y= -(x^2- 3)= 3- x^2. Again, we swap x and y and solve for y. x= 3- y^2] so y^2= 3- x and y= \pm\sqrt{3- x}. Because, here, x< 0 becomes y< 0, we must take the negative sign. y= -\sqrt{3- x}.

    If 0< x\le \sqrt{3}, we have the same situation as the previous case except that y is positive so we take the positive sign on the square root: y= \sqrt{3- x}.

    If \sqrt{3}< x, we have the same situation as the first case except that y is positive so we take the positive sign on the square root: y= \sqrt{x- 3}.
    Last edited by HallsofIvy; November 15th 2012 at 06:24 AM.
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