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

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Answer 1 · 2015-02-20T10:35:11

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

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

so:

$x^{2} - 2 x \geq 0 \Rightarrow x (x - 2) \geq 0$ $x^2-2x>=0rArrx(x-2)>=0$ ,

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

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

so:

$x \leq 0 \lor x \geq 2$ $x<=0vvx>=2$ , or, in another notation,

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

How do you find values of x where the function f(x)=√x2−2xf(x)=sqrt(x^2 - 2x) is continuous?