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The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on
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Verified Answer
The correct answer is:
$(0, \infty)$
Given, $f(x)=\log (1+x)-\frac{2 x}{2+x}$
$$
\begin{aligned}
\therefore \quad f^{\prime}(x) &=\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \\
&=\frac{x^{2}}{(1+x)(x+2)^{2}}
\end{aligned}
$$
Clearly, $f^{\prime}(x)>0$ for all $x>0$. Hence, $f(x)$ is increasing on $(0, \infty)$.
$$
\begin{aligned}
\therefore \quad f^{\prime}(x) &=\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \\
&=\frac{x^{2}}{(1+x)(x+2)^{2}}
\end{aligned}
$$
Clearly, $f^{\prime}(x)>0$ for all $x>0$. Hence, $f(x)$ is increasing on $(0, \infty)$.
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