Question

Show that f(x) = $\frac{1}{1+x^2}$ is neither increasing nor decreasing on R.

Answer 1

$$\begin{gathered} f\left(x\right)=\frac{1}{1+x^2} \\ R\text{can be divided into two intervals}\left(0,\infty \right)\text{and}(-\infty ,0]. \\ \mathrm{Case}1:\text{Let}{x}_1,x_2\in \left(0,\infty \right)\text{such that}{x}_1<x_2.\text{Then,} \\ {x}_1<x_2 \\ \Rightarrow {x_1}^2<{x_2}^2 \\ \Rightarrow 1+{x_1}^2<1+{x_2}^2 \\ \Rightarrow \frac{1}{1+{x_1}^2}>\frac{1}{1+{x_2}^2} \\ \Rightarrow f\left(x_1\right)>f\left(x_2\right)\forall x_1,x_2\in \left(0,\infty \right) \\ \text{So,}f\left(x\right)\text{is decreasing on}\left(0,\infty \right)\text{.} \\ \mathrm{Case}2:\text{Let}{x}_1,x_2\in (-\infty ,0]\text{such that}{x}_1<x_2.\text{Then,} \\ {x}_1<x_2 \\ \Rightarrow {x_1}^2>{x_2}^2 \\ \Rightarrow 1+{x_1}^2>1+{x_2}^2 \\ \Rightarrow \frac{1}{1+{x_1}^2}<\frac{1}{1+{x_2}^2} \\ \Rightarrow f\left(x_1\right)<f\left(x_2\right)\forall x_1,x_2\in (-\infty ,0] \\ \text{So,}f\left(x\right)\text{is increasing on}(-\infty ,0]\text{.} \\ \text{Here,}f\left(x\right)\text{is decreasing on}\left(0,\infty \right)\text{and}\mathrm{increasing}\text{on}(-\infty ,0]\text{.} \\ \text{Thus,}f\left(x\right)\text{is neither increasing nor decreasing on}R. \end{gathered}$$