OpenStudy (mendicant_bias):

I'm assuming it's still the case, but say you have a function with two terms involving x; if one ends up undefined (div. by zero e.g.), but the other doesn't, the function's output is still considered undefined as a whole for that input, right?

4 years ago
OpenStudy (anonymous):

a halo fan eh? i guess you mean: \[f(x) = g(x) + h(x)\] where g(x) is undefined, but h(x) is? then yeah, you're right.

4 years ago
OpenStudy (mendicant_bias):

Yeah. Thanks! And, yes, lol.

4 years ago
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