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.
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The function you have proposed is continuous everywhere. But you do have the right idea. Try this. $\displaystyle f(x) = \left\{ {\begin{array}{rr} { - 1} & {x < 0} \\ 1 & {0 \le x} \\ \end{array}} \right.$
Yeah I meant x + 1 if x does not equal zero -1 if x equals zero.
Well yes. That will work also.
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