# What is the range of the function f(x)=(3x^2+3x-6)/(x^2-x-12)?

Dec 30, 2017

The range is $y \in \left(- \infty , 0.614\right] \cup \left[2.692 , + \infty\right)$

#### Explanation:

Let $y = \frac{3 {x}^{2} + 3 x - 6}{{x}^{2} - x - 12}$

To find the range, proceed as follows

$y \left({x}^{2} - x - 12\right) = 3 {x}^{2} + 3 x - 6$

$y {x}^{2} - 3 {x}^{2} - y x - 3 x - 12 y + 6 = 0$

${x}^{2} \left(y - 3\right) - x \left(y + 3\right) - \left(12 y - 6\right) = 0$

This is a quadratic equation in $x$ and in order for this equation to have solutions, the discriminant $\Delta \ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(- \left(y + 3\right)\right)}^{2} - 4 \left(y - 3\right) \left(- \left(12 y - 6\right)\right) \ge 0$

${y}^{2} + 6 y + 9 + 4 \left(y - 3\right) \left(12 y - 6\right) \ge 0$

${y}^{2} + 6 y + 9 + 4 \left(12 {y}^{2} - 42 y + 18\right) \ge 0$

${y}^{2} + 6 y + 9 + 48 {y}^{2} - 168 y + 72 \ge 0$

$49 {y}^{2} - 162 y + 81 \ge 0$

$y = \frac{162 \pm \sqrt{{162}^{2} - 4 \cdot 49 \cdot 81}}{2 \cdot 49}$

$= \frac{162 \pm 101.8}{98}$

Therefore,

The range is $y \in \left(- \infty , 0.614\right] \cup \left[2.692 , + \infty\right)$

graph{(3x^2+3x-6)/(x^2-x-12) [-14.24, 14.23, -7.12, 7.12]}

Dec 30, 2017

Range: $f \left(x\right) \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{3 {x}^{2} + 3 x - 6}{{x}^{2} - x - 12}$ or

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

$f \left(x\right) = 0$ for $\left(x = 1 , x = - 2\right)$

$f \left(x\right)$ is undefined for $\left(x = - 3 , x = 4\right)$

$f \left(x\right) = \infty \mathmr{and} f \left(x\right) = - \infty$ when $x$ approaches $- 3 \mathmr{and} 4$

Therefore range is any real value ,i.e$f \left(x\right) \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

Range: $f \left(x\right) \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

graph{(3x^2+3x-6)/(x^2-x-12) [-40, 40, -20, 20]} [Ans]