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Thread: Range of Rational Functions

  1. #1
    Senior Member harpazo's Avatar
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    Range of Rational Functions

    Find the range of f(x) = (2x^2)/(x^4 + 1).

    How is this done algebraically?
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    MHF Contributor MarkFL's Avatar
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    Re: Range of Rational Functions

    I would write:

    $\displaystyle y=\frac{2x^2}{x^4+1}$

    And then arrange as:

    $\displaystyle yx^4-2x^2+y=0$

    This is a quadratic in $\displaystyle x^2$, and so we require the discriminant to be non-negative:

    $\displaystyle (-2)^2-4(y)(y)\ge0$

    $\displaystyle 4(1-y^2)\ge0$

    $\displaystyle (y+1)(y-1)\le0$

    And from this we conclude:

    $\displaystyle -1\le y\le1$

    Next, we look at the roots of the quadratic:

    $\displaystyle x^2=\frac{-(-2)2\pm\sqrt{(-2)^2-4(y)(y)}}{2(y)}=\frac{1\pm\sqrt{1-y^2}}{y}$

    We see the numerator cannot be negative, so we only require the denominator to be positive (since a square cannot be negative or undefined), and combined with the restriction on the discriminant, we then conclude:

    $\displaystyle 0<f(x)\le1$
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Range of Rational Functions

    That should be:

    $\displaystyle 0\le f(x)\le1$

    when we look at $\displaystyle x=0$.
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  4. #4
    Senior Member harpazo's Avatar
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    Re: Range of Rational Functions

    Quote Originally Posted by MarkFL View Post
    I would write:

    $\displaystyle y=\frac{2x^2}{x^4+1}$

    And then arrange as:

    $\displaystyle yx^4-2x^2+y=0$

    This is a quadratic in $\displaystyle x^2$, and so we require the discriminant to be non-negative:

    $\displaystyle (-2)^2-4(y)(y)\ge0$

    $\displaystyle 4(1-y^2)\ge0$

    $\displaystyle (y+1)(y-1)\le0$

    And from this we conclude:

    $\displaystyle -1\le y\le1$

    Next, we look at the roots of the quadratic:

    $\displaystyle x^2=\frac{-(-2)2\pm\sqrt{(-2)^2-4(y)(y)}}{2(y)}=\frac{1\pm\sqrt{1-y^2}}{y}$

    We see the numerator cannot be negative, so we only require the denominator to be positive (since a square cannot be negative or undefined), and combined with the restriction on the discriminant, we then conclude:

    $\displaystyle 0<f(x)\le1$
    Nicely done. Good study notes.
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