Question
Without using the derivative, show that the function f(x) = |x| is
- Strictly increasing in $(0,\infty)$
- Strictly decreasing in $(-\infty,0)$
✓
Answer
We have,
$\text{f}(\text{x})=|\text{x}|=\begin{cases}\text{x},&\text{x}>0\\\text{-x},&\text{x}<0\end{cases}$
So, f(x) is increasing in $(0,\infty).$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
So, f(x) is decreasing on $(-\infty,0).$
$\text{f}(\text{x})=|\text{x}|=\begin{cases}\text{x},&\text{x}>0\\\text{-x},&\text{x}<0\end{cases}$
- Let $\text{x}_1,\text{x}_2\in(0,\infty)$ and $\text{x}_1>\text{x}_2$
So, f(x) is increasing in $(0,\infty).$
- Let $\text{x}_1,\text{x}_2\in (-\infty,0)$ and $\text{x}_1>\text{x}_2$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
So, f(x) is decreasing on $(-\infty,0).$
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