# Question 757f2

Feb 24, 2017

the domain is all the real vales of x, or: (−∞,∞)
the range is: y≥1 or [1,∞)#

#### Explanation:

$f \left(x\right) = \sqrt{{x}^{2} + 1}$
The domain:
A square root function is defined if where the entire statement under the radical sign is equal or greater than 0, so in this case:
${x}^{2} + 1 \ge 0$ => or:
${x}^{2} \ge - 1$=> this is always true, hence the domain is all the real vales of $x$, or: $\left(- \infty , \infty\right)$
The range:
In its valid domain the absolute minimum value of $f \left(x\right)$ occurs when $x = 0$, therefore the range is:
$y \ge 1$ or $\left[1 , \infty\right)$