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

## $\mathrm{Here},\phantom{\rule{0ex}{0ex}}f\left(x\right)=\left|x\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right)\text{Let}{x}_{1},{x}_{2}\in \left(0,\infty \right)\text{such that}{x}_{1}<{x}_{2}.\text{Then,}\phantom{\rule{0ex}{0ex}}{x}_{1}<{x}_{2}\phantom{\rule{0ex}{0ex}}⇒\left|{x}_{1}\right|<\left|{x}_{2}\right|\phantom{\rule{0ex}{0ex}}⇒f\left({x}_{1}\right)\left|{x}_{2}\right|\phantom{\rule{0ex}{0ex}}⇒f\left({x}_{1}\right)>f\left({x}_{2}\right)\phantom{\rule{0ex}{0ex}}\text{∴}{x}_{1}<{x}_{2}⇒f\left({x}_{1}\right)>f\left({x}_{2}\right),\forall {x}_{1},{x}_{2}\text{∈}\left(-\infty ,0\right]\text{.}\phantom{\rule{0ex}{0ex}}\text{So,}f\left(x\right)\text{is decreasing on}\left(-\infty ,0\right]\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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