The function f(x)=log (1+x)-frac{2 x}{2+x} is | vedclass

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(D) Given the function $f(x) = \log(1+x) - \frac{2x}{2+x}$. The domain of the function requires $1+x > 0$,so $x > -1$.First,we find the derivative $f'

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$(-1, \infty)$

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(D) Given the function $f(x) = \log(1+x) - \frac{2x}{2+x}$. The domain of the function requires $1+x > 0$,so $x > -1$.First,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx} [\log(1+x)] - \frac{d}{dx} \left[ \frac{2x}{2+x} \right]$$f'(x) = \frac{1}{1+x} - \left[ \frac{(2+x)(2) - (2x)(1)}{(2+x)^2} \right]$$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$Simplifying the expression:$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$For the function to be increasing,we require $f'(x) > 0$.Since $x^2 > 0$ and $(2+x)^2 > 0$ for all $x > -1$ (excluding $x=0$),the condition $f'(x) > 0$ is satisfied when $1+x > 0$,which implies $x > -1$.Therefore,the function is increasing on $(-1, \infty)$.

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

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