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Yes. Note that the function in your example is not defined for f(x) < 0. The domain is then x >= 0.
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Wow. So wrong. Much sadness.
Counterexample: f(x) = 1
OP, the domain of a function can be almost whatever you want it to be.
For example, f(x)=x and domain = {1,2,3}.
Then for all values of x != 1,2, or 3, my function is not defined.
Note, that if I actually plugged in a value for x, I would still get an output, but that output is not an output of my function as defined.
It is common in some applications to ignore irrelevant parts of the domain for your functions. For example, position of a thrown baseball after time 0. There is no negative time, so our relevant domain is D = {x|x>=0}.
There are also functions that have potential inputs that would result in no outputs, and so their domains
must exclude those inputs.
For example, f(x) = 1/x. This function cannot have an input of 0, and therefore its domain cannot have 0 in it.
OP, back to the counterexample I started this post with, and relevant to your second-to-last post, a function like f(x) = e^x will result in f(x) > 0 for all x. Same with f(x) = e^(-x). Just like f(x) = 1 in that f(x) > 0 for all x.