# What is the domain of h(x)=sqrt(x^2 - 2x + 5)?

Sep 6, 2015

Domain: $\left(- \infty , + \infty\right)$

#### Explanation:

Since you're dealing with the square root of an expression, you know that you need to exclude from the domain of the function any value of $x$ that will make the expression under the square root negative.

For real numbers, the square root can only be taken from positive numbers, which means that you need

${x}^{2} - 2 x + 5 \ge 0$

Now you need to find the values of $x$ for which the above inequality is satisfied. Look what happens when you use a little algebraic manipulation to rewrite the inequality

${x}^{2} - 2 x + 5 \ge 0$

${x}^{2} - 2 x + 1 + 4 \ge 0$

${\left(x - 1\right)}^{2} + 4 \ge 0$

Because ${\left(x - 1\right)}^{2} \ge 0$ for any value of $x \in \mathbb{R}$, it follows that

${\left(x - 1\right)}^{2} + 4 \ge 0 \text{, } \left(\forall\right) x \in \mathbb{R}$

This means that the domain of the function can include all real numbers, since you cannot have a negative expression under the square root regardless of which $x$ you plug in.

In interval notation, the domain of the function will thus be $\left(- \infty , + \infty\right)$.

graph{sqrt(x^2-2x+5) [-10, 10, -5, 5]}