Show that f(x) = 4 - 3x is one-one. Show that g(x) = x² + 6x is not one-one on ℝ…

Step 1. (a). $4 - 3a = 4 - 3b$ forces $a = b$.

Step 2. (b). $g(0) = 0$ and $g(-6) = 36 - 36 = 0$: two inputs, one output.

Step 3. (c). $g(x) = (x + 3)^2 - 9$ turns at $x = -3$. On $[-3, \infty)$ it only rises, so it is one-one there.