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What are your thoughts?
For what values of x do you think you might have to worry about a discontinuity?
I haven't learned about discontinuity, I don't know why this is on my quiz.
oh my. Ok.... well, it's probably in the lesson somewhere, but that's ok. For THESE purposes just think of continuity like this: a function is CONTINUOUS if I can draw its graph without lifting my pencil from the paper. If there is a place where I have to lift my pencil, then it is "discontinuous" at that point.
Does that much make sense?
Now, what you have here is a piecewise function - do you understand what that means? It is actually defined by a linear equation on part of its domain (I already told you that's going to be continuous), and a quadratic equation on the other part of its domain (and as I said, that too is continuous). so all you have to worry about is the point where the equation "changes" from the linear equation, to the quadratic.... that's the only x where there MIGHT be a discontinuity.
so the questions are: 1. what is that x value, where you have to check for continuity? and 2. at that x-value, do the 2 "pieces" of the functions "meet up"? E.g., if they meet at that point, then the function is continuous there - even though the equation changes, you don't have to lift the pencil! :) if the function "jumps" at that point, then it isn't continuous there.
that^^^ is just an example by the way, it isn't YOUR function.
Alright, if I have understood correctly and worked this correctly, it is discontinuous.
Going further with that, I would choose A as my answer.
Is that correct? @DebbieG
Well done. :) it is discontinuous at x=5, because of the "jump" in the graph.
Thank you very much.
f(5)=0 then the endpoint of the other "segment" of the graph is at the point (5,15) (but it's an open circle,not actually ON the graph).
notice that if the defintion for x>5 had said: f(x)=x^2-25 for x>5 Then it would not longer be discontinuous, it would be continuous everywhere. :) and, you're welcome!