3

I have a question on this sheet of lecture notes here, answering a question asking to find these functions and state their domains: enter image description here

My query lies with the bottom of the page. Consider the following:

$(g\ o\ f)(x) = 3 -\frac{6}{x}$

Apparently, this cannot equal 2. In that case the output is 0. How is this a domain restriction? It's known, and at 0.

sangstar
  • 1,947
  • The function $h(x) = \frac xx$ is also restricted, as $x$ cannot be $0$. However, we can simplify the expression describing $h$ to $h(x) = 1$. Can you insert $x = 0$ into that expression? Certainly. Is $h$ still restricted? Yes, it is. Changing the expression used to describe the function does not change the function itself in any way. – Arthur Jun 14 '17 at 07:26

1 Answers1

2

Strictly speaking, $(g\ o\ f)(x)$ is a composite of function by plugging in arbitrary values of $x$ into $f$ and then taking the value $f(x)$ into the function $g$.

Since function $f$ cannot admit the value $x=2$, then it is clear that $(g\ o\ f)(x)$ cannot admit the value $x=2$, even though yes there is a value obtained when you substitute $x=2$ directly into the final expression.

Soby
  • 1,482