Hi folks,

I really hope there is no simple arithmetic error in my answer, but I've checked it many times, so here goes.

Show that for real x the function $y = \frac{x + 2}{(x + 3)(x - 1)}$ can take all real values.

This should be really easy:

$y = \frac{x + 2}{x^2 + 2x - 3}$

$yx^2 + 2yx - 3y = x + 2 $

$yx^2 + (2y - 1)x - 3y - 2 = 0$

a quadratic in x of the form $ax^2 + bx + c = 0$. For real x the discriminant $(b^2 - 4ac \ge 0)$

$(2y - 1)^2 + 4.y.(3y + 2) \ge 0$

$4y^2 - 4y + 1 + 12y^2 + 8y \ge 0$

$16y^2 + 4y + 1 \ge 0$

solving for y, this has no real roots. I was expecting to get any expression like $y^2 + something \ge 0$ which would be true for all y, but no real roots suggests I have taken a wrong turn. Can you see it.