Let f:(-1, infty) rightarrow mathbb{R} be def | vedclass

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(A) For $x 
eq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} ight) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x

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decreases in $(-1, \infty)$

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(A) For $x 
eq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} ight) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)}$.Let $h(x) = x - (1+x)\ln(1+x)$.Then $h'(x) = 1 - [\ln(1+x) + (1+x) \cdot \frac{1}{1+x}] = 1 - \ln(1+x) - 1 = -\ln(1+x)$.For $x \in (-1, 0)$,$1+x \in (0, 1)$,so $\ln(1+x)  0$.For $x \in (0, \infty)$,$1+x > 1$,so $\ln(1+x) > 0$,which implies $h'(x) Since $h(0) = 0 - (1)\ln(1) = 0$,$h(x)$ increases on $(-1, 0)$ and decreases on $(0, \infty)$.Thus,$h(x) Since $x^2(1+x) > 0$ for all $x \in (-1, \infty) \setminus \{0\}$,$f'(x) = \frac{h(x)}{x^2(1+x)} Therefore,the function $f$ is strictly decreasing on $(-1, \infty)$.

Let $f:(-1, \infty) \rightarrow \mathbb{R}$ be defined by $f(0)=1$ and $f(x)=\frac{1}{x} \ln(1+x), x \neq 0$. Then the function $f$

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