Question

Without using the derivative, show that the function f (x) = | x | is
(a) strictly increasing in (0, ∞)
(b) strictly decreasing in (−∞, 0).

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ f\left(x\right)=\left|x\right| \\ \left(\mathrm{a}\right)\text{Let}{x}_1,x_2\in \left(0,\infty \right)\text{such that}{x}_1<x_2.\text{Then,} \\ {x}_1<x_2 \\ \Rightarrow \left|x_1\right|<\left|x_2\right| \\ \Rightarrow f\left(x_1\right)<f\left(x_2\right) \\ \text{∴}{x}_1<x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right),\forall x_1,x_2\in \left(0,\infty \right) \\ \text{So,}f\left(x\right)\text{is increasing on}\left(0,\infty \right)\text{.} \\ \left(\mathrm{b}\right)\text{Let}{x}_1,x_2\in (-\infty ,0]\text{.}\text{such that}{x}_1<x_2.\text{Then,} \\ {x}_1<x_2 \\ \Rightarrow \left|x_1\right|>\left|x_2\right| \\ \Rightarrow f\left(x_1\right)>f\left(x_2\right) \\ \text{∴}{x}_1<x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right),\forall x_1,x_2\text{∈}(-\infty ,0]\text{.} \\ \text{So,}f\left(x\right)\text{is decreasing on}(-\infty ,0]\text{.} \end{gathered}$$