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Math Help - Is my inverse right?

  1. #1
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    Is my inverse right?

    I had to find the inverse of a function, and it sort of feels right but looks wrong if you know what I mean.

    I was given

    Is my inverse right?-001.png

    and came up with

    Is my inverse right?-002.png

    Don't feel the x should be under the root somehow. Any ideas?
    Attached Thumbnails Attached Thumbnails Is my inverse right?-001.png   Is my inverse right?-001.png  
    Last edited by alexpasty2013; January 7th 2014 at 10:52 AM.
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  2. #2
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    Re: Is my inverse right?

    No, that is not correct. For one thing, in the original function, f(x)= -3(x-2)^2+ 12, when x= 4, y= 0. But with your function, f^{-1}(x)= 1+ \sqrt{\frac{x- 12}{3}}, does not have a real number when x= 0. Also since the original function has limits on x, so should the inverse function.

    As for x being "under the root", since in the original function x is squared, I can't see why you should expect the "opposite" of that, the square root of x, for the inverse function. But how did you get x- 12, rather than 12- x?
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  3. #3
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    Re: Is my inverse right?

    Quote Originally Posted by HallsofIvy View Post
    No, that is not correct. For one thing, in the original function, f(x)= -3(x-2)^2+ 12, when x= 4, y= 0. But with your function, f^{-1}(x)= 1+ \sqrt{\frac{x- 12}{3}}, does not have a real number when x= 0. Also since the original function has limits on x, so should the inverse function.

    As for x being "under the root", since in the original function x is squared, I can't see why you should expect the "opposite" of that, the square root of x, for the inverse function. But how did you get x- 12, rather than 12- x?
    The domain/range (x axis) is restricted to being less than or equal to 1 or less than or equal so as to eliminate the possibility of two roots. Also an unrestricted function f(x)=-3(x-1)^2+12 has y=9 when x is 0 and x= -1 and x=3 when y = 0. So not sure where you are getting x=4 from?

    I got x-12 by subtracting 12 from both sides. I then put the whole of both sides over -3 to get rid of the -3 which multiplies the brackets on the right, then put the both sides under a root to un-square the bracket. Tried replacing (x-1)^2 with (x-1)(x-1) and multiplying out and got into a bit of a pickle.
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  4. #4
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    Re: Is my inverse right?

    This is how I do it:

    y = -3(x - 1)^2 + 12

    12 - y = 3(x - 1)^2

    4 - \frac{y}{3} = (x - 1)^2

    Now here we have a problem....we don't know which square root to take, yet. More on that later.

    \pm \sqrt{4 - \frac{y}{3}} = x - 1

    1 \pm \sqrt{4 - \frac{y}{3}} = x

    now when x is 0, y is 9, so this helps us decide which square root we need:

    1 + \sqrt{4 - \frac{9}{3}} = 1 + \sqrt{4 - 3} = 1 + 1 = 2. Wrong one.

    1 - \sqrt{4 - \frac{9}{3}} = 1 - \sqrt{4 - 3} = 1 - 1 = 0. Bingo!

    So it ought to be that:

    f^{-1}(x) = 1 - \sqrt{4 - \frac{x}{3}}.

    Just as a safeguard, let's see if f \circ f^{-1}(x) = x:

    f \circ f^{-1}(x) = f\left(1 - \sqrt{4 - \frac{x}{3}}\right) = -3\left(1 - \sqrt{4 - \frac{x}{3}} - 1\right)^2 + 12

    = -3\left(-\sqrt{4 - \frac{x}{3}}\right)^2 + 12 = -3\left(4 - \frac{x}{3}\right) + 12 = -12 + x + 12 = x
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  5. #5
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    Re: Is my inverse right?

    Of course, deveno's 4- \frac{x}{3} is the same as \frac{12- x}{3}. But note it is 12- x, not x- 12.
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  6. #6
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    Re: Is my inverse right?

    Quote Originally Posted by alexpasty2013 View Post
    The domain/range (x axis) is restricted to being less than or equal to 1 or less than or equal so as to eliminate the possibility of two roots. Also an unrestricted function f(x)=-3(x-1)^2+12 has y=9 when x is 0 and x= -1 and x=3 when y = 0. So not sure where you are getting x=4 from?

    I got x-12 by subtracting 12 from both sides. I then put the whole of both sides over -3 to get rid of the -3 which multiplies the brackets on the right,
    There's your difficulty. Dividing x- 12 by -3 gives [tex]\frac{x- 12}{-3}= \frac{12- x}{3}. You don't have "-3" in the denominator so you must have 12- x, not x- 12, in the numerator

    then put the both sides under a root to un-square the bracket. Tried replacing (x-1)^2 with (x-1)(x-1) and multiplying out and got into a bit of a pickle.
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