# How do you find values of x where the function f(x)=sqrt(x^2 - 2x) is continuous?

Feb 20, 2015

The answer is: $\left(- \infty , 0\right] \bigcup \left[2 , + \infty\right)$.

The domain of a square root (and of all the roots with even index) like $\sqrt{f} \left(x\right)$, is $f \left(x\right) \ge 0$,

so:

${x}^{2} - 2 x \ge 0 \Rightarrow x \left(x - 2\right) \ge 0$,

considering that $0$ and $2$ are the solutions of the equation:

$x \left(x - 2\right) = 0$ and considering that the inequality is $\ge$, than the solutions are for external values,

so:

$x \le 0 \vee x \ge 2$, or, in another notation,

$\left(- \infty , 0\right] \bigcup \left[2 , + \infty\right)$.