Rules that are and aren't functions [f(x) = (x-3)/(x-3)]

Hi,

I am reviewing my maths and came across definition of a function. According to it, for each input(x) it needs to have corresponding output(y). This one - f(x) = (x-3)/(x-3) - doesn't have y for x=3. The same for y=sqrt(x) or y=5x^2 + x^1/4 - first one doesn't have y for x<0 and the second one for x>0 has two values.

Despite of this they(or similar expressions) are called in some books or videos as 'functions'.

My question is does it matter if they are formally called functions or not if we actually can do with them all the things we do with functions?

I would appreciate if someone could explain it to me :)

Re: Rules that are and aren't functions [f(x) = (x-3)/(x-3)]

First, a "rule", by itself, is never a function. One way of defining a functions is, simply "a set of ordered pairs such that no pairs have the same first member with different second members". If we are given both a domain (the set of first members) and a "rule" (so that, for each first member we can calculate the second member) then we have a function.

Though it is not good notation, we are sometimes given a "rule" along with the (unstated!) assumption that the domain is the largest set of values to which the "rule" can be applied. That is, if we are given the rule "f(x)= (x-3)/(x- 3)", the default domain is "all x except 3". But it is still a function. Similarly, $\displaystyle f(x)= \sqrt{x}$, is a function **with domain** "all non-negative x".

The most important of your examples is "y=5x^2 + x^1/4". If it were true that, as you say, "for x>0 [it] has two values", then it would NOT be a function. However, you are wrong that y, or x^1/4, "has two values". Yes, it is true that $\displaystyle y^4= x$ has two roots. But x^1/4 is, by definition, the **positive** root of y. The equatilon $\displaystyle x^4= a$ has (real number) solutions $\displaystyle x= \pm a^{1/4}$. The reason we need to write "$\displaystyle \pm$" is that "$\displaystyle a^{1/4}$", by itself, means only one of those two solutions.

(in terms of complex numbers, there would be **four** solutions to $\displaystyle x^4= a$.)