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

Nov 10, 2015

Domain : $\mathbb{R}$.
Range : [2,+oo[.

#### Explanation:

The domain of $f$ is the set of real $x$ such that ${x}^{2} - 2 x + 5 \ge 0$.

You write ${x}^{2} - 2 x + 5 = {\left(x - 1\right)}^{2} + 4$ (canonical form), so you can see that ${x}^{2} - 2 x + 5 > 0$ for all real $x$. Therefore, the domain of $f$ is $\mathbb{R}$.

The range is the set of all values of $f$. Because $x \mapsto \sqrt{x}$ is an increasing function, the variations of $f$ are same than $x \mapsto {\left(x - 1\right)}^{2} + 4$ :
- $f$ is increasing on [1,+oo[,
- $f$ is decreasing on ]-oo,1].
The minimal value of $f$ is $f \left(1\right) = \sqrt{4} = 2$, and f has no maximum.

Finally, the range of $f$ is [2,+oo[.