Hi guys, this is part of the harder portion of my homework, and I'm struggling to know where to start. Could someone point me in the right direction? Thanks a lot.
a) In this exercise, we find the range (i.e. the possible values) of (6x + 5) / (3x2 + 4x + 2)
Let D = (6x + 5) / (3x2 + 4x + 2) for all x.
i) Show that 3Dx2 + (4D - 6)x + 2D - 5 = 0
ii) Hence, show that 3Dx2 + (4D - 6)x + 2D - 5 = 0 has real solutions in x if and only if (D - 3)(2D + 3) <= 0
iii) Hence, find the range of (6x + 5) / (3x2 + 4x + 2)
b) Find the range of (x2 - 6x + 5) / (x2 + 2x + 1)
Notice that you get a quadratic term (which can be factorised) inside that square root. If you did your calculations correctly, it should read inside the square root. You can get a real root if an only if this is a non-negative quantity, because you can't take (real) square roots of negative numbers.