Find the Domain and Range square root of x^2+1

$\sqrt{x^{2} + 1}$

Step 1

Set the radicand in $\sqrt{x^{2} + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} + 1 \geq 0$

Step 2

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$x^{2} \geq - 1$

Since the left side has an even power, it is always positive for all real numbers.

All real numbers

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${y | y \geq 1}$

Step 5

Determine the domain and range.

Domain: $(- \infty, \infty), {x | x \in ℝ}$

Range: $[1, \infty), {y | y \geq 1}$

Step 6

Find the Domain and Range square root of x^2+1

Related Questions

Support LCMGCF.com by disabling your adblocker.