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

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Answer 1 · 2015-02-20T10:35:11

The answer is: $(-oo,0] uuu [2,+oo)$ .

The domain of a square root (and of all the roots with even index) like $sqrtf(x)$ , is $f(x)>=0$ ,

so:

$x^2-2x>=0rArrx(x-2)>=0$ ,

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

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

so:

$x<=0vvx>=2$ , or, in another notation,

$(-oo,0]uuu[2,+oo)$ .

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