# Find the domain and range of the following functions interval notation f(x)=x^2+4x and g(x)=6/(1-x) ?

Feb 2, 2017

$\left(1\right) : f : \mathbb{R} \to \left[- 4 , \infty\right) .$

$\left(2\right) : g : \left(- \infty , 1\right) \cup \left(1 , \infty\right) \to \left(- \infty , 0\right) \cup \left(0 , \infty\right) .$

#### Explanation:

We will discuss the Soln. in $\mathbb{R} .$

$f \left(x\right) = {x}^{2} + 4 x .$

So, to operate $f$, we can take choose any $x$ from $\mathbb{R}$,

meaning that, the Domain is $\mathbb{R}$.

Next, we note that, $f \left(x\right) = {x}^{2} + 4 x = \left({x}^{2} + 4 x + 4\right) - 4 = {\left(x + 2\right)}^{2} - 4.$

Also, AA x in RR, (x+2)^2>=0;" adding "-4, (x+2)^2-4>=-4.

$\Rightarrow f \left(x\right) \ge - 4$

$\therefore \text{ The Range is } \left[- 4 , \infty\right) .$

Regarding, $g \left(x\right) = \frac{6}{1 - x} , \text{ we can not divide by 0; so, } \left(1 - x\right) \ne 0.$

In other words, the Domain of

$g \text{ is } \mathbb{R} - \left\{1\right\} = \left(- \infty , 1\right) \cup \left(1 , \infty\right)$.

Further, since, $g \left(x\right) = 6 \left(\frac{1}{1 - x}\right) , \mathmr{and} , \forall x \in \mathbb{R} - \left\{1\right\} , \frac{1}{1 - x} \ne 0$

$\Rightarrow g \left(x\right) \ne 0.$

$\Rightarrow \text{ the Range of g is } \mathbb{R} - \left\{0\right\} = \left(- \infty , 0\right) \cup \left(0 , \infty\right) .$

Enjoy Maths.!