I'm pretty sure this statement is false. Let f be a defined function. If abs(f) is continuous at a, then f is also continuous at a. Is this an accetable counterexample? f(x) = x + 1 if x does not equal zero 1 if x equals zero.
Follow Math Help Forum on Facebook and Google+
The function you have proposed is continuous everywhere. But you do have the right idea. Try this.
Yeah I meant x + 1 if x does not equal zero -1 if x equals zero.
Well yes. That will work also.
View Tag Cloud