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

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Application of Derivatives

Solution:

$f'\left(x\right)=\frac{1}{1+x}-\frac{\left(2+x\right)2-2x-1}{\left(2+x\right)^{2}}$
$=\frac{1}{1+x}-\frac{4}{\left(2+x\right)^{2}}$
$=\frac{4+x^{2}+4x-4-4x}{\left(1+x\right)\left(2+x\right)^{2}}$
$=\frac{x^{2}}{\left(1+x\right)\left(2+x\right)^{2}} > 0$
[if $1 + x > 0$ i.e., if $x - 1$]