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Question
Show that f(x) = \[\frac{1}{1 + x^2}\] decreases in the interval [0, ∞) and increases in the interval (−∞, 0] ?
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Solution
\[\text { Here }, \]
\[f\left( x \right) = \frac{1}{1 + x^2}\]
\[\text { Case 1: Let} \text{x}_1 , x_2 \in \left( 0, \infty \right) \text { such that } x_1 < x_2 . \text { Then },\]
\[ x_1 < x_2 \]
\[ \Rightarrow {x_1}^2 < {x_2}^2 \]
\[ \Rightarrow 1 + {x_1}^2 < 1 + {x_2}^2 \]
\[ \Rightarrow \frac{1}{1 + {x_1}^2} > \frac{1}{1 + {x_2}^2}\]
\[ \Rightarrow f\left( x_1 \right) > f\left( x_2 \right) \forall x_1 , x_2 \in \left( 0, \infty \right)\]
\[\text { So, }f\left( x \right) \text { is decreasing on }\left( 0, \infty \right).\]
\[\text { Case } 2: Let x_1 , x_2 \in ( - \infty , 0]\text { such that } x_1 < x_2 . \text { Then, }\]
\[ x_1 < x_2 \]
\[ \Rightarrow {x_1}^2 > {x_2}^2 \]
\[ \Rightarrow 1 + {x_1}^2 > 1 + {x_2}^2 \]
\[ \Rightarrow \frac{1}{1 + {x_1}^2} < \frac{1}{1 + {x_2}^2}\]
\[ \Rightarrow f\left( x_1 \right) < f\left( x_2 \right)\]
\[ \Rightarrow f\left( x_1 \right) < f\left( x_2 \right), \forall x_1 , x_2 \in ( - \infty , 0]\]
\[\text { So, }f\left( x \right) \text { is increasing on } ( - \infty , 0].\]
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