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Question
Without using the derivative, show that the function f (x) = | x | is.
(a) strictly increasing in (0, ∞)
(b) strictly decreasing in (−∞, 0) .
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Solution
\[\text { Here }, \]
\[f\left( x \right) = \left| x \right|\]
\[(a) \text { Let } x_1 , x_2 \in \left( 0, \infty \right) \text { such that } x_1 < x_2 . \text { Then },\]
\[ x_1 < x_2 \]
\[ \Rightarrow \left| x_1 \right| < \left| x_2 \right|\]
\[ \Rightarrow f\left( x_1 \right) < f\left( x_2 \right)\]
\[\therefore x_1 < x_2 \Rightarrow f\left( x_1 \right) < f\left( x_2 \right), \forall x_1 , x_2 \in \left( 0, \infty \right)\]
\[\text { So },f\left( x \right) \text { is increasing on }\left( 0, \infty \right).\]
\[(b) \text { Let } x_1 , x_2 \in ( - \infty , 0]. \text { such that } x_1 < x_2 . \text { Then },\]
\[ x_1 < x_2 \]
\[ \Rightarrow \left| x_1 \right| > \left| x_2 \right|\]
\[ \Rightarrow f\left( x_1 \right) > f\left( x_2 \right)\]
\[\therefore x_1 < x_2 \Rightarrow f\left( x_1 \right) > f\left( x_2 \right), \forall x_1 , x_2 \in ( - \infty , 0].\]
\[\text { So },f\left( x \right) \text { is decreasing on }( - \infty , 0].\]
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