Discuss the Continuity of the function on the closed interval (Calculus)?

It just wants me to like talk about the continuity of the interval for example....

g(x) = squarert (25-x^2) ; and with an interval of [-5,5]
The answer is Continuous on [-5,5]

But the next question is f(t) = 3 - sqrt(9-t^2); and with an interval of [-3,3]

Could someone explain this to me? I looked in the back for the solution to the first one and I'm still not really getting it. Like is it continuous for everywhere but 0? That's what I guessed it was but can someone kind of explain it to me please. Thanks.
Yeah thanks Andrew that really did help. I got it now. I just did x/(x^2-1) and figured that there was a non removable discontinuity at x=1 since 1 would make the denominator 0. So yeah I got it now. Thanks.
 
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