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If there exists a value for which y = undefined on the graph, it is discontinuous.

For example: y = x/(x+2) would be discontinuous at x = -2 as this would be x/0 (which is undefined)

2006-09-12 12:11:37 · answer #1 · answered by p_rutherford2003 5 · 0 0

(I assume here that you are referring to functions defined from the reals to the reals; things are significantly more complicated if you want to generalize)

If you have a function f(t), it is continuous if for every epsilon, there exists some delta such that if

| t - c | < delta

then

| f(t) - f(c) | < epsilon

In other words, small "changes in t" result in small "changes in f."

This is a rigorous definition. A less rigorous definition is to say that all continuous functions can be drawn on a chalkboard without ever lifting the piece of chalk from the board.

If you want to prove that a function is continuous, you usually can follow a procedure like this...

*) Take some epsilon > 0 and some t

*) Assign a value of delta that depends on epsilon (you'll have to figure out which value to assign here; this is the trick)

*) Take some c with t - delta < c < t + delta.

*) Show that such a c causes |f(c) - f(t)| < epsilon

If you can do that, then your function is continuous.

2006-09-12 12:33:13 · answer #2 · answered by Ted 4 · 0 0

If the derivative and nth derivatives are continuous throughout its domain.

2006-09-12 13:50:03 · answer #3 · answered by Anonymous · 0 0

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