The function f(x)= x 1+|x| is: 1. Stritcly increasing. 2. Stritcl… - Vidyadip

Question

The function $\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$ is:

Stritcly increasing.
Stritcly decreasing.
Neither increasing nor decreasing.
None of these.

✓

Answer

Stritcly increasing.

Solution:

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

Case I:

When x > 0, |x| = x

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

$=\frac{\text{x}}{1+\text{x}}$

$\Rightarrow\text{f}'(\text{x})=\frac{(1+\text{x})1-\text{x}(1)}{(1+\text{x})^2}$

$=\frac{1}{(1+\text{x})^2}>0,\forall\ \text{x}\in\text{R}$

So, f(x) is strictly increasing when x > 0.

Case II:

When x < 0, |x| = -x

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

$=\frac{\text{x}}{1+\text{x}}$

$\Rightarrow\text{f}'(\text{x})=\frac{(1-\text{x})1-\text{x}(-1)}{(1-\text{x})^2}$

$=\frac{1}{(1-\text{x})^2}>0,\forall\ \text{x}\in\text{R}$

So, f(x) is strictly increasing when x < 0.

Thus, f(x) is strictly increasing on R.

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