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

Question

8934 views around the world

You can reuse this answer
Creative Commons License

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)#.