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