Question
Show that the function $f(x)=\frac{x-2}{x+1}, x \neq-1$ is increasing.

Answer

$
\begin{aligned}
& f(x)=\frac{x-2}{x+1} \\
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x-2}{x+1}\right)=\frac{(x+1) \cdot \frac{d}{d x}(x-2)-(x-2) \cdot \frac{d}{d x}(x+1)}{(x+1)^2}
\end{aligned}
$
$
\begin{gathered}
\quad=\frac{(x+1) \cdot(1-0)-(x-2) \cdot(1+0)}{(x+1)^2} \\
=\frac{x+1-x+2}{(x+1)^2}=\frac{3}{(x+1)^2} \\
\because x \neq-1 \quad \therefore x+1 \neq 0 \\
\therefore(x+1)^2>0 \quad \therefore \frac{3}{(x+1)^2}>0
\end{gathered}
$
$\therefore f ^{\prime}( x )>0$, for all $x \in R , x \neq-1$
Hence, the function $f$ is increasing for all $x \in R$, where $x \neq-1$.

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