In which interval is the function f(x) = log | vedclass

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(B) To find the interval where the function $f(x) = \log x - \frac{2x}{2 + x}$ is increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx

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$(0, \infty)$

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(B) To find the interval where the function $f(x) = \log x - \frac{2x}{2 + x}$ is increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(\log x) - \frac{d}{dx}\left(\frac{2x}{2 + x}\right)$Using the quotient rule for the second term: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$f'(x) = \frac{1}{x} - \frac{(2)(2 + x) - (2x)(1)}{(2 + x)^2}$$f'(x) = \frac{1}{x} - \frac{4 + 2x - 2x}{(2 + x)^2}$$f'(x) = \frac{1}{x} - \frac{4}{(2 + x)^2}$$f'(x) = \frac{(2 + x)^2 - 4x}{x(2 + x)^2}$$f'(x) = \frac{4 + x^2 + 4x - 4x}{x(2 + x)^2} = \frac{x^2 + 4}{x(2 + x)^2}$Since $x^2 + 4 > 0$ and $(2 + x)^2 > 0$ for all $x$ in the domain of $\log x$ (which is $x > 0$),the sign of $f'(x)$ depends only on $x$.For $f(x)$ to be increasing,we require $f'(x) > 0$.$\frac{x^2 + 4}{x(2 + x)^2} > 0 \implies x > 0$.Thus,the function is increasing in the interval $(0, \infty)$.

In which interval is the function $f(x) = \log x - \frac{2x}{2 + x}$ an increasing function?

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