# How do you find the domain and range of f(x)= (2-x)/(x^2+7x+12)?

Jan 28, 2017

The domain is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 4 , - 3\right\}$
The range is  ]-oo, -21.95]uu[-0.455,+oo[

#### Explanation:

Let's factorise the denominator

${x}^{2} + 7 x + 12 = \left(x + 4\right) \left(x + 3\right)$

As you cannot divide by $O$, $x \ne - 4$ and $x \ne - 3$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 4 , - 3\right\}$

Let $y = \frac{2 - x}{{x}^{2} + 7 x + 12}$

Then,

$y {x}^{2} + 7 y x + 12 y = 2 - x$

$y {x}^{2} + 7 y x + x + 12 y - 2 = 0$

$y {x}^{2} + \left(7 y + 1\right) x + 12 y - 2$

Solving for $x$

The discriminant is

$\Delta = {b}^{2} - 4 a c$

$= {\left(7 y + 1\right)}^{2} - 4 \cdot y \cdot \left(12 y - 2\right)$

$= 49 {y}^{2} + 14 y + 1 - 48 {y}^{2} + 8 y$

$= {y}^{2} + 22 y + 1$

This has to be $\ge 0$

Therefore,

${22}^{2} - 4 \cdot 1 = 484 - 4 = 480$

So,

$y = \frac{- 22 \pm \sqrt{480}}{2}$

$= - 11 \pm 2 \sqrt{30}$

Therefore,

y in ]-oo, -21.95]uu[-0.455,+oo[

The range is f(x) in ]-oo, -21.95]uu[-0.455,+oo[