# How do you find the domain and range of  sqrt(x^2 - 8x +15)?

Mar 4, 2018

Domain: $x \in \left(- \infty , 3\right] \cup \left[4 , \infty\right)$

Range: $y \in {\mathbb{R}}_{\ge 0}$

#### Explanation:

The domain of a function is the intervals where the function is defined in terms of real numbers.

In this case we have a square root, and if we have negative numbers under a square root, the expression will be undefined, so we need to solve for when the expression under the square root is negative. This is the same as solving the inequality:

${x}^{2} - 8 x + 15 < 0$

Quadratic inequalities are easier to work out if we factor them, so we factor by grouping:

${x}^{2} - 3 x - 5 x + 15 < 0$

$x \left(x - 3\right) - 5 \left(x - 3\right) < 0$

$\left(x - 5\right) \left(x - 3\right) < 0$

In order for the expression to be negative, only one of the factors may be negative (mind you, a negative times a negative is a positive and a positive times a positive is a positive). We can see that the only time this happens is on the interval $x \in \left(3 , 5\right)$

This means that we need to exclude $\left(3 , 5\right)$ from our domain, which gives a domain of $\left(- \infty , 3\right] \cup \left[5 , \infty\right)$

The possible resulting values of a square root is all positive values and zero, and since the bit inside the square root is continuous and spans all the necessary values, we know that the range must be all positive real numbers and zero, ${\mathbb{R}}_{\ge 0}$