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Math Help - Finding Removable Discontinuities

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    Finding Removable Discontinuities

    I have the function (x^2+3x)/(x^3-x), and I have to find it's removable discontinuity. I factored it, and got x(x+3)/x(x+1)(x-1). The single x in the top and the bottom cancels out, and solving for zero obviously gives me x=0 for the removable discontinuity. However, whenever I graph the function, I consistently find that the graph is undefined at 0,-3...
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    Re: Finding Removable Discontinuities

    Quote Originally Posted by qubez View Post
    I have the function (x^2+3x)/(x^3-x), and I have to find it's removable discontinuity. I factored it, and got x(x+3)/x(x+1)(x-1). The single x in the top and the bottom cancels out, and solving for zero obviously gives me x=0 for the removable discontinuity. However, whenever I graph the function, I consistently find that the graph is undefined at 0,-3...
    You must be entering incorrectly.
    Attached Thumbnails Attached Thumbnails Finding Removable Discontinuities-untitled.gif  
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    Re: Finding Removable Discontinuities

    Is there just no discontinuity? There isn't one at x=0, is there?
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    Re: Finding Removable Discontinuities

    Quote Originally Posted by qubez View Post
    Is there just no discontinuity? There isn't one at x=0, is there?
    Well there are three discontinuities: -1,~0,~1.
    But 0 is a removable discontinuity; \pm 1 are essential discontinuities.
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    Re: Finding Removable Discontinuities

    I guess I'm just confused because there are asymptotes at x=-1 and x=1, so the graph never gets near x=0, so how could you insert a point at x=0 and make the graph continuous? That question probably made no sense...
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    Re: Finding Removable Discontinuities

    Quote Originally Posted by qubez View Post
    I guess I'm just confused because there are asymptotes at x=-1 and x=1, so the graph never gets near x=0, so how could you insert a point at x=0 and make the graph continuous? That question probably made no sense...
    Are we looking at the same graph? Redefine the function (let's call it f) so that f(0)=-3. Then it's continuous at x=0.
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    Re: Finding Removable Discontinuities

    Quote Originally Posted by Reckoner View Post
    Are we looking at the same graph? Redefine the function (let's call it f) so that f(0)=-3. Then it's continuous at x=0.
    I must ask you, so what is new in your reply?
    Yes x=0 is removable as I said. SO what is new?
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