S.O.S. Mathematics CyberBoard

[sno_more]

Let me set this up, ask my question, then explain why I am confused or confirm my suspicion.

Let g(x)= (4-x)^1/2 i.e., g(x) equals the square root of (4 minus x). The domain of g(x) is all x less than or equal to four.

Let f(x)= 1/x (i.e., the reciprocity function). The domain of f(x) is all x, other than 0.

It would seem to me that the domain of the composite, f(g(x)), would exclude x=4, but not exclude x=0. I get to this result because if g(4)=0, that feeds f(x) as f(0) or 1 divided by zero, which is undefined. So, in the composite, x=4 results in division by zero. On the other hand, 0 would be in the domain of the composite, because for the composite, if x=0, then g(0)=2, so that 1/g(0)= 1/2, which is a perfectly permissible and pedestrian outcome. Am I right so far?

This conclusion works, assuming the domain of g(x) can properly be defined so that it results in no values in the range of g(x) such that, when the values of the range of g(x) are defined as the domain of f(x), f(x) operates to, for example, divide by zero. A value that, in the domain of g(x), corresponds to a range value for g(x) which is in the domain of f(x) is in the domain of the composite, even though it is outside the domain values of f(x).

I think this is important to the proof that the composite of continuous functions is itself continuous (perhaps not to others, but to me). But I recall from somewhere that one must exclude from the domain of a composite function any value which is excluded from the domain of any of the functions of which it is composed. But perhaps I am misremembering that, and that is only true as an element of the relationship between reciprocal functions?

[Kwyjibo]

By definition the domain of f(g(x)) is the set of all values of x for which g(x) is defined and f(g(x)) is defined.

-Kwyjibo
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[sno_more]

Thanks.

The polite way to describe your response is blunt. Another observation might be that to use a definition as the answer assumes the conclusion.

Can you get there without the resort to definition? I haven't been able to.

[Mr Fantastic]

Nutshell: The range of g has to be a subset of the domain of f in order that f[g(x)] is defined.
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Ciao baby!

[Kwyjibo]

[sno_more]

Thanks. To both of you. For both the long and the short versions of the answer, both very helpful. In a colloquial nutshell, I shouldn't assume the negative pregnant.

Now, for another question, if I may trouble you. Is it a characteristic of a cusp that the right and left hand limits of the first derivative at the point of the cusp approach positive and negative infinity, respectively (which being which depending on whether the cusp is a local maximum or minimum of the function)? If so, why? This blew out of nowhere on me, as the explanation in the back of a book for the difference between a cusp and a corner of a continuous piecewise defined function. I like it, but I don't understand it.

[Mr Fantastic]

[qleak]