variation table of $h=g\circ f$ Let 
$h(x)=\dfrac{6x^2+8x+11}{(x-1)^2}$ and $h=g\circ f$ such that
 $f(x)=\dfrac{2x+3}{x-1}$ and $g(x)=x^2+2$
Deduce   variation table of $h$  at $(-\infty,\frac{-3}{2}]$, $[\frac{-3}{2},1)$ and  $(1,+\infty)$
Indeed i found the variation table of $f$ and $g$
\begin{array}{c|cc|} 
 x &      -\infty & 1 &  +\infty \\ \hline
 f(x) &    \searrow & || & \searrow 
 \end{array}
\begin{array}{c|cc|} 
 x &      -\infty & 0 &  +\infty \\ \hline
 h(x) &    \searrow & 2 & \nearrow
 \end{array}
i would like to use $h=g\circ f$ but my problem i don't know to get the image of 
$f((-\infty,1))$ and $f((1,+\infty))$ 
i know that $$x< 1 \implies f(x)>f(1) $$ but $f(1)$ can't be determinate 
the same for $$x> 1 \implies f(x)<f(1) $$
note that the student not yet know the notion of limits and continuity 

Write $f$ as $f(x)=2+5/(x-1)$ and let's analyse the term $1/(x-1)$ at $(1,\infty)$. If we start with $x=2$ for example we have $1/(x-1)=1$ and if we take $x=3,4,5,...$ we see that $1/(x-1)=1/2, 1/3, 1/4, ...$. Can you guess what is happening?. The value of $1/(x-1)$ is decreasing and as bigger as you choose $x$, the term $1/(x-1)$ is getting smaller and smaller until reaches $0$. If you take a very, very big value for $x$, let's say $\infty$, the value of $f$ will became $2$ because the term $5/(x-1)$ vanishes. Now let's see what happen when $x$ gets close to $1$. If we take the numbers $x=1+1/2, 1+1/3, 1+1/4, 1+1/5,...$, the term $1/(x-1)$ will become $2, 3, 4, 5,...$. See what is happening now? It is increasing as you wish. So if you get very close to $1$ the term $1/(x-1)$ will be very, very big, let's say $\infty$, and then we conclude that $f((1,\infty))=(2,\infty)$. If you make the same analysis for $(-\infty,1)$ you will get the image $(-\infty,2)$.
