Question

state the domain and range.

1. $f(x) = \frac{2x(x - 4)}{(x - 6)(x + 3)}$

domain
range
$\frac{-4\div2}{-18}$ $\frac{2}{9}$

Explanation:

Step1: Find domain (denominator ≠ 0)

Set denominator $(x-6)(x+3)
eq 0$. Solve $x-6=0 \implies x=6$; $x+3=0 \implies x=-3$. So domain excludes $x=-3,6$.

Step2: Define domain in interval form

Domain: $(-\infty, -3) \cup (-3, 6) \cup (6, \infty)$

Step3: Simplify function for range analysis

$f(x)=\frac{2x(x-4)}{(x-6)(x+3)}=\frac{2x^2-8x}{x^2-3x-18}$. Let $y=\frac{2x^2-8x}{x^2-3x-18}$, rearrange to $(y-2)x^2 + (-3y+8)x -18y=0$

Step4: Use discriminant for real x

For real $x$, discriminant $\Delta \geq 0$. $\Delta=(-3y+8)^2 -4(y-2)(-18y) \geq 0$
Expand: $9y^2-48y+64 +72y^2-144y \geq 0 \implies 81y^2-192y+64 \geq 0$

Step5: Solve quadratic inequality

Factor: $(9y-8)^2 \geq 0$. This is true for all real $y$, since squares are non-negative.

Answer:

Domain: $\boldsymbol{(-\infty, -3) \cup (-3, 6) \cup (6, \infty)}$
Range: $\boldsymbol{(-\infty, \infty)}$ (all real numbers)