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

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(A) Given,$f(x) = \log(1+x) - \frac{2x}{2+x}$.First,we find the derivative $f'(x)$ with respect to $x$:$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)

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$(0, \infty)$

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(A) Given,$f(x) = \log(1+x) - \frac{2x}{2+x}$.First,we find the derivative $f'(x)$ with respect to $x$:$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$$f'(x) = \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2}$$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$$f'(x) = \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 $x^2 \geq 0$ and $(2+x)^2 > 0$ for $x > -1$ (domain of $\log(1+x)$),the sign of $f'(x)$ depends on $(1+x)$.Thus,$f'(x) > 0$ when $1+x > 0$,which means $x > -1$.However,checking the options provided,$f'(x) > 0$ for all $x > 0$ is clearly satisfied.Therefore,the function is increasing on $(0, \infty)$.

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

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