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

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(A) To determine where $f(x)$ is increasing,we find its derivative $f'(x)$.Given $f(x) = \log(1+x) - \frac{2x}{2+x}$.The domain of $f(x)$ is $x > -1$.

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$(-1, \infty)$

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(A) To determine where $f(x)$ is increasing,we find its derivative $f'(x)$.Given $f(x) = \log(1+x) - \frac{2x}{2+x}$.The domain of $f(x)$ is $x > -1$.$f'(x) = \frac{d}{dx} [\log(1+x)] - \frac{d}{dx} [\frac{2x}{2+x}]$.Using the quotient rule for the second term: $\frac{d}{dx} [\frac{2x}{2+x}] = \frac{(2+x)(2) - 2x(1)}{(2+x)^2} = \frac{4+2x-2x}{(2+x)^2} = \frac{4}{(2+x)^2}$.So,$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$.$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+4x+x^2-4-4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$.For $f(x)$ to be increasing,we require $f'(x) > 0$.Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x$ in the domain,$f'(x) > 0$ whenever $1+x > 0$,which means $x > -1$.Thus,$f(x)$ is increasing in $(-1, \infty)$.

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

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