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

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(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.Differentiating the function with resp

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$(-1, \infty)$

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(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.Differentiating the function with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{(2+x)(1) - x(1)}{(2+x)^2} \right]$$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{2+x-x}{(2+x)^2} \right] = \frac{1}{1+x} - \frac{4}{(2+x)^2}$$f^{\prime}(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+x^2+4x-4-4x}{(1+x)(2+x)^2}$$f^{\prime}(x) = \frac{x^2}{(1+x)(2+x)^2}$.Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x > -1$,the sign of $f^{\prime}(x)$ depends on the term $\frac{1}{1+x}$.For $f(x)$ to be increasing,we require $f^{\prime}(x) > 0$.Since $x > -1$,$1+x > 0$,which implies $f^{\prime}(x) > 0$ for all $x \in (-1, \infty)$.Thus,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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