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  4. The function f(x) = log(1+ x ) - (2x/2+x) is increasing on

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Q. The function $f\left(x\right) = log\left(1+ x \right) - \frac{2x}{2+x}$ is increasing on

Application of Derivatives

Solution:

Given, $f\left(x\right) = log\left(1+ x \right) - \frac{2x}{2+x}$
$\Rightarrow f'\left(x\right) = \frac{1}{1+x} - \frac{\left(2+x\right)2-2x}{\left(2+x\right)^{2}}$
$\Rightarrow f'\left(x\right) = \frac{x^{2}}{\left(x+1\right)\left(x+2\right)^{2}}$
Clearly, $f'\left(x\right) > 0$ for all $x > -1$, hence $f\left(x\right)$ is increasing on $\left(-1, \infty\right)$.