Composing function The function is $f: \Bbb{R}\rightarrow\Bbb{R}$ defined as $f(x)= 2/ (x -3).$
I need to find $(f o f)(1).$
I would like to ask which of the following answers are the right one for writing this function.
$( f o f) ( 1 ) = ( f ( f ( 1 ) ) )= ( f ( 2/ 1 - 3) = ( 2 / 1 - 3 - 2 ) = -1/2 $
or
$( f o f ) ( 1 ) = ( f ( f (1) ) )= f( 2/ 1 - 3 ) = 2 / ( 2/ 1 - 3 ) - 3 = 2 / ( - 2 ) - 3 = -4$
 A: None of them, you know that $f(1)=-1$ and $(f \circ f)(1)=f(f(1))$ so you substitute $x$ with $f(1)$ and you have $(f \circ f)(1)=\frac{2}{(-1)-3}=\frac{1}{-2}$.
A: If $f(x) = \frac{2}{x} -3 $, then
\begin{align*}
(f\circ f)(x) = f(f(x)) = \frac{2}{f(x)}-3 = \frac{2}{\frac{2}{x}-3}-3
\end{align*}
And thus
\begin{align*}
(f\circ f)(1) = \frac{2}{\frac{2}{1}-3}-3 = \frac{2}{-1}-3 = -5
\end{align*}
A: It's easier if you simplify the expressing each step so that $$f \circ f(1)=f(f(1))=f(2/(1-3))=f(-1)=2/(-1- 3)=-\frac12$$
and so we see your first calculation is correct.
A: At first note that we should consider a different domain of the function $f$:
\begin{align*}
&f:\mathbb{R}\setminus\{3\}\to\mathbb{R}\\
&f(x)=\frac{2}{x-3}
\end{align*}
since $f$ is not defined at $x=3$.
We can calculate $(f\circ f)(1)=f(f(1))$ as follows:
\begin{align*}
\color{blue}{f(f(1))}=
\begin{cases}
\frac{2}{f(1)-3}=\frac{2}{\frac{2}{1-3}-3}=\frac{2}{-1-3}\color{blue}{=-\frac{1}{2}}\\
f\left(\frac{2}{1-3}\right)=f(-1)=\frac{2}{-1-3}\color{blue}{=-\frac{1}{2}}
\end{cases}
\end{align*}
