Show that f(x)=1 1+x^2 is neither increasing nor decreasing on R. - Vidyadip

Question

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

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Answer

Here, $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ R can be divided into two intervals $0,\infty$ and $(-\infty,0].0,$Case 1:
Let $\text{x}_1,\text{x}_2\in (0,\infty)$ such that $\text{x}_1<\text{x}_2.$ Then,$\text{x}_1<\text{x}_2$
$\Rightarrow\text{x}_1^2<\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2<1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}>\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ So, f(x) is decreasing on $(0,\infty).$Case 2:
Let $\text{x}_1,\text{x}_2\in(-\infty,0]$ such that $\text{x}_1<\text{x}_2.$Then, $\text{x}_1<\text{x}_2$
$\Rightarrow\text{x}_1^2>\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2>1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}<\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(-\infty,0]$ So, f(x) is increasing on $(-\infty,0].$ Here, f(x) is decreasing on $(0,\infty)$ and increasing on $(-\infty,0].$ Thus, f(x) is neither increasing nor decreasing on R.

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