# How do you find domain and range for f(x) =sqrt (x^2 - 2x + 5)?

Oct 4, 2015

Complete the square to find that the domain of $f \left(x\right)$ is the whole of $\mathbb{R}$ and its range is $\left[2 , \infty\right)$

#### Explanation:

$f \left(x\right) = \sqrt{{x}^{2} - 2 x + 5} = \sqrt{{\left(x - 1\right)}^{2} + 4}$

${\left(x - 1\right)}^{2} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$

So $f \left(x\right) = \sqrt{{\left(x - 1\right)}^{2} + 4}$ is defined for all $x \in \mathbb{R}$

So the (implicit) domain of $f \left(x\right)$ is $\mathbb{R}$.

$f \left(x\right)$ has minimum value when $\left(x - 1\right) = 0$, that is when $x = 1$.

$f \left(1\right) = \sqrt{{0}^{2} + 4} = \sqrt{4} = 2$

So the range of $f \left(x\right)$ is $\left[2 , \infty\right)$