0

Assume $g(x)=x$ and $f(x)=1/x$. Then, $f(x)$ is not defined for $x=0$. Is $f(x)\times g(x)$ defined for every $x$? The intuition says no because multiplication of functions is equal to the value of one function multiplied by the value of the other for every $x$, and if we cant calculate the value of $g(x)$ for $x=0$ then we can't multiply this value by $g(x)$. But if I will replace the function by their equation I will get that $f(x)\times g(x)=1$ for every $x$.

ultralegend5385
  • 10,106
daniel
  • 141
  • 3
    Please format your post using MathJax – AllCatsAreGrey Aug 09 '24 at 12:59
  • $x/x=1$ is only valid if $x\ne 0$. – Karl Aug 09 '24 at 13:01
  • 1
    In Mathematics case matters. So $X$ and $x$ are not the same. $F$ and $f$ are not the same. – jjagmath Aug 09 '24 at 13:04
  • It's always ok to simplify functions. Of course, simplifying a function means computing its domain and, for all $x$ in the domain, an alternative way to write the value of the function. – Sassatelli Giulio Aug 09 '24 at 13:05
  • Technically, you're not allowed to say $\frac1x\cdot x=1$ "for every $x$" because of the trouble at $x=0$. So, the graph of $f(x)\cdot g(x)$ would have a hole at $x=0$. However, precisely because the algebraic expression simplifies, the limit as $x$ approaches $0$ is well-defined (namely $1$), so that the trouble amounts to what we call a "removable discontinuity": the limit fills the hole. For the sake of expediency, especially w/trig simplifications, we often implicitly treat removable discontinuities as having been removed. It wouldn't hurt to be clearer about when and why we do this. – Blue Aug 09 '24 at 13:12

1 Answers1

3

You keep all restrictions.

If $f:X\subseteq \mathbb{R}\to \mathbb{R}$ and $g:Y\subseteq \mathbb{R}\to \mathbb{R}$ are two functions with domains $X$ and $Y$ respectively, the function $f\cdot g$ is only defined on the intersection $X\cap Y$ independent of whether $f\cdot g$ can naturally be extended to a larger domain.

Related but not entirely the same; If someone gives you the function defined by $f(x)=\frac{(x-1)x^2}{(x-1)}$, the largest meaningful domain is $\mathbb{R}\setminus \{1\}$ as you cannot divide by zero. That means that $f$ is different from the function $g(x)=x^2$ although $f(x)=g(x)$ for all $x\neq 1$.

Mathematician 42
  • 13,176
  • I have upvoted this answer, but its final sentence has left out an assumption, namely that we are dealing with a function $g:{\mathbf R}\rightarrow{\mathbf R}$. That the domain of $g$ should be the set of real numbers is a natural and typically widespread default, but remains an assumption. – Paul Tanenbaum Aug 09 '24 at 13:27
  • @PaulTanenbaum : Indeed, the same goes for $f$ in the second to last line. I don't like the practice of defining functions by only specifying the functional equation, though it is common in calculus (or settings where we're tipcally dealing with real functions). One should never forget that the domain of a function is an integral part of the definition. – Mathematician 42 Aug 09 '24 at 15:47
  • Music to my ears, Mathematician 42. Even decades later, I am still distressed that my advisor, whom I respect tremendously, believes it not only acceptable but appropriate in certain introductory instructional contexts to identify a function with its “rule”… essentially as a collection of ordered pairs, rather than as a binary relation $f \subsetneq A \times B$ for specific $A$ and $B$ (and, obviously, with the existence and uniqueness properties). – Paul Tanenbaum Aug 09 '24 at 16:35