Question
Show that $\text{f}(\text{x})=\tan^{-1}\text{x}-\text{x}$ is a decreasing function on R.

Answer

$\text{f}(\text{x})=\tan^{-1}\text{x}-\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2}-1$
$=\frac{1-1-\text{x}^2}{1+\text{x}^2}$
$=\frac{-\text{x}^2}{1+\text{x}^2}$
We know,
$\text{x}^2\geq0,1+\text{x}^2>0,\forall\ \text{x}\in\text{R}$
$\therefore\ \frac{-\text{x}^2}{1+\text{x}^2}<0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in\text{R}$
Hence, f(x) is decreasing on R.

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