Question

Show that f(x) = tan⁻¹ x − x is a decreasing function on R.

Answer 1

$$\begin{gathered} f\left(x\right)={\tan }^{-1}x-x \\ f\text{'}\left(x\right)=\frac{1}{1+x^2}-1 \\ =\frac{1-1-x^2}{1+x^2} \\ =\frac{-x^2}{1+x^2} \\ \text{We know,} \\ {x}^2\ge 0,\text{1+}{\mathit{\text{x}}}^2\text{>0, ∀}x\in \mathrm{R} \\ \therefore \frac{-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}<0,\forall x\in \mathrm{R} \\ \Rightarrow f\mathit{\text{'}}\left(\mathrm{x}\right)<0,\forall x\in \mathrm{R} \\ \text{So,}f\left(x\right)\mathit{}\mathrm{is}\mathrm{decreasing}\mathrm{on}R. \end{gathered}$$