What is the domain of the function: f(x) =sqrt(x^2 - 2x + 5)?

Sep 17, 2015

${D}_{f} = R$

Explanation:

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

$D = {b}^{2} - 4 a c = {\left(- 2\right)}^{2} - 4 \cdot 1 \cdot 5 = 4 - 20 = - 16$

Because $D < 0$ and $a = 1 > 0$, expression ${x}^{2} - 2 x + 5 > 0$ for $\forall x \in R$ and square root can be calculated. Hence,

${D}_{f} = R$