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Showing that a function can take any real values if x is real. watch

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    Show that ,if x is real , the function 2-x /(x^2-4x+1) can take any real value.

    Hello if anyone can help me prove that, i am stuck.
    Im getting negative roots but it says x is real....
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    Set y=f(x) rearrange into a quadratic then find the discriminant which will be a function of y which you can then show is always greater than 0.
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    (Original post by black1blade)
    Set y=f(x) rearrange into a quadratic then find the discriminant which will be a function of y which you can then show is always greater than 0.
    Yes i did all that, but my discriminant is turning to negative.
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    (Original post by Bilbao)
    Yes i did all that, but my discriminant is turning to negative.
    The question is not entirely correct because the discriminant of x^2-4x+1 =0 is 4^2-4 = 12 > 0 therefore there are two points in \mathbb{R} for which this quadratic is 0 and hence your function is undefined.

    Anyway, ignoring those two points, this is rather missing the actual question, because it wants you to show that y takes all the values in \mathbb{R} rather than x. To show this, first find any horizontal asymptotes and show that this graph does indeed cross them. Secondly, show that that you have y \rightarrow \pm \infty at some point point(s) along it.
 
 
 
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