If log (1+x)-frac{2x}{2+x} is increasing,then | vedclass

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(A) Let $f(x) = \log(1+x) - \frac{2x}{2+x}$.For the function to be defined,we must have $1+x > 0$,which implies $x > -1$.Now,find the derivative $f'(x

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$0 < x < \infty$

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(A) Let $f(x) = \log(1+x) - \frac{2x}{2+x}$.For the function to be defined,we must have $1+x > 0$,which implies $x > -1$.Now,find the derivative $f'(x)$:$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$$f'(x) = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$For the function to be increasing,$f'(x) > 0$:$\frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} > 0$$\frac{4+x^2+4x - 4 - 4x}{(1+x)(2+x)^2} > 0$$\frac{x^2}{(1+x)(2+x)^2} > 0$Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for $x 
eq -2$,the inequality holds when $1+x > 0$ and $x 
eq 0$.Thus,$x > -1$ and $x 
eq 0$.

If $\log (1+x)-\frac{2x}{2+x}$ is increasing,then

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