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(C) To find the interval where the function $f(x) = 2x + \log \left(\frac{x}{2+x}\right)$ is increasing,we first find its derivative $f'(x)$.First,not

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$(-\infty, -2) \cup (0, \infty)$

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(C) To find the interval where the function $f(x) = 2x + \log \left(\frac{x}{2+x}ight)$ is increasing,we first find its derivative $f'(x)$.First,note that the domain requires $\frac{x}{2+x} > 0$,which implies $x \in (-\infty, -2) \cup (0, \infty)$.$f(x) = 2x + \log(x) - \log(2+x)$.Differentiating with respect to $x$:$f'(x) = 2 + \frac{1}{x} - \frac{1}{2+x}$.$f'(x) = 2 + \frac{2+x-x}{x(2+x)} = 2 + \frac{2}{x(2+x)} = \frac{2x(2+x) + 2}{x(2+x)} = \frac{2(x^2 + 2x + 1)}{x(2+x)} = \frac{2(x+1)^2}{x(2+x)}$.For the function to be increasing,we require $f'(x) > 0$.Since $2(x+1)^2 \ge 0$ for all $x$,$f'(x) > 0$ when $x(2+x) > 0$ and $x 
eq -1$.The inequality $x(2+x) > 0$ holds for $x \in (-\infty, -2) \cup (0, \infty)$.Thus,the function is increasing in the interval $(-\infty, -2) \cup (0, \infty)$.

The interval in which the curve represented by $f(x) = 2x + \log \left(\frac{x}{2+x}\right)$ is increasing is

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