1. Range of Rational Functions

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

How is this done algebraically?

2. 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:

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

3. Re: Range of Rational Functions

That should be:

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

when we look at $\displaystyle x=0$.

4. Re: Range of Rational Functions

Originally Posted by MarkFL
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.