# Let  f(x)= (x+3)/(x^2 -4x-12) and g(x)=sqrt(3-x), how do you find each of the compositions and domain and range?

Sep 15, 2015

As explained below.

#### Explanation:

$f \circ g \left(x\right) =$$\frac{\sqrt{3 - x} + 3}{\left(3 - x\right) - 4 \sqrt{3 - x} - 12}$

$g \circ f \left(x\right) =$sqrt(3- (x+3)/(x^2-4x-12)

f(x) can be written as $\frac{x + 3}{\left(x - 6\right) \left(x + 2\right)}$,its domain would therefore be $\left(- \infty , - 2\right) U \left(- 2 , 6\right) U \left(6 , \infty\right)$, or $\left\{x \in \mathbb{R} , x \ne - 2 , 6\right\}$

For its range interchange x and y and solve the resultant quadratic equation for y. The result would be

y=$\frac{\left(4 x + 1\right) \pm \sqrt{\left(4 x + 1\right) \left(16 x + 1\right)}}{2 x}$

This would mean $x \le - \frac{1}{4}$ or $x \ge - \frac{1}{16}$
Range would be$\left\{y : \mathbb{R} , y \le - \frac{1}{4} \mathmr{and} y \ge - \frac{1}{16}\right\}$

For g(x), domain would be $\left\{x : \mathbb{R} , x \le 3\right\}$ so that all y values are real.

Range would be $\left\{y : \mathbb{R} , y \ge 0\right\}$