Question

Show that the function $f\left(x\right)=\frac{x-2}{x+1}$, for $x\ne 0$ is increasing.

Answer 1

$\begin{array}{cc}f\left(x\right)=\frac{x-2}{x+1}& f\left(x\right)>0\\ f^{\prime }\left(x\right)=\frac{\left(x+1\right)\left(1\right)-\left(x-2\right)\left(1\right)}{{\left(x+1\right)}^2}& \\ =\frac{1+2}{{\left(x+1\right)}^2}=\frac{3}{{\left(x+1\right)}^2}& \\ f^{\prime }\left(x\right)isalways+ve& \end{array}$

hence $f\left(x\right)$ is increasing.