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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions5 Marks
Question
Show that $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ is decreases in the interval $[0,\infty)$ and increases in the interval $(-\infty,0].$
✓

Answer

Here, $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$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 $\text{x}_1,\text{x}_2\in (0,\infty]$ such that $\text{x}_1<\text{x}_2.$ Then, $\text{x}_1<\text{x}_2$ $\Rightarrow\text{x}_1^2>\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2<1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}<\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(0,\infty]$ So, f(x) is increasing on $(0,\infty].$

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