Question

f(x) =
$\frac{1}{ \sqrt{x + 2} }$
what is Domain and range​

Answer 1

Answer:

$\sf \green{ To \: Find \: Domain :-}$

$\displaystyle \sf \green{f(x) = \frac{1}{ \sqrt{x + 2} } }$

Seperate the function into parts to determine the domain of each part.

$\displaystyle \sf \green{ \frac{1}{ \sqrt{x + 2} } }$

$\displaystyle \sf \green{ \sqrt{x + 2} }$

$\displaystyle \sf \green{x + 2}$

The domain for rational function are all values of x for which denominator is different than 0.

$x \in \mathbb{R} \lbrace-2 \rbrace$

$\displaystyle \sf \green{x \geqslant - 2 }$

$\sf \: x \in \mathbb{R}$

Find intersection.

$\displaystyle \sf \green{x \in \: ⟨- 2 \: , + \infty \rang}$

$\sf \therefore \: Domain :- x \in ⟨-2 , +∞ ⟩$

$Range :R^{+} - \lbrace \: 0 \rbrace$