Is f(x) = sqrt(x) continuous on interval [1, 2]?

[karan]

Is funtion f(x) = sqrt(x) ,where x belongs to interval [1,2] continious?

 

[stapel]

Is funtion f(x) = sqrt(x) ,where x belongs to interval [1,2] continious?

What did you conclude when you looked at the graph?

Eliz.

 

[pka]

Perhaps I am just too pessimistic, but I can’t see this post as serious.
Can any of you see anyone asking this question not seeing at once the answer?

 

[stapel]

Perhaps I am just too pessimistic, but I can’t see this post as serious.
Can any of you see anyone asking this question not seeing at once the answer?

This person has posted twenty-two questions, of varying difficulty and from disparate areas of mathematics, in about a day's time. No work has been shown on any of them (even the ones for which answers had already been provided to his posts elsewhere), nor has he ever responded to any replies or queries.

This could just be a lazy-arse cheater, or this could be one of those psychology-class assignments, where the student is told to see how obnoxious he can be before he starts ticking people off. Dunno.

Eliz.

 

[galactus]

You are so right, Stapel and pka. That's why I deleted my post. I know the problem is rather obvious, but there are plenty like that yet still they want a proof of some sort. This is one of those posters who posts a litany of problems and then never comes back to say thanks, follow up, or anything at all. They post then forget about it.