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

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(A) Given the function $f(x) = \log(1 + x) - \frac{2x}{2 + x}$.To find the interval where the function is increasing,we calculate the derivative $f'(x

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$(0, \infty)$

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(A) Given the function $f(x) = \log(1 + x) - \frac{2x}{2 + x}$.To find the interval where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx} [\log(1 + x)] - \frac{d}{dx} \left[ \frac{2x}{2 + x} \right]$Using the quotient rule for the second term:$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}$Simplify the expression:$f'(x) = \frac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2}$$f'(x) = \frac{4 + 4x + x^2 - 4 - 4x}{(1 + x)(2 + x)^2}$$f'(x) = \frac{x^2}{(1 + x)(2 + x)^2}$For the function to be increasing,we require $f'(x) > 0$.Since $x^2 \ge 0$ and $(2 + x)^2 > 0$ for all $x > -1$ (domain of $\log(1+x)$),the sign of $f'(x)$ depends on $(1 + x)$.$f'(x) > 0$ when $1 + x > 0$,which means $x > -1$.However,looking at the options provided,$f'(x) > 0$ for all $x > 0$ is clearly satisfied.Thus,$f(x)$ is increasing on $(0, \infty)$.

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

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