Let $$f : (-1, \infty) \to R$$ be defined by $$f(0) = 1$$ and $$f(x) = \frac{1}{x}\log_e(1 + x)$$, $$x \ne 0$$. Then the function $$f$$:

We are asked to study the monotonicity of the function

$$f : (-1,\infty)\;\longrightarrow\; \mathbb R,\qquad f(x)=\begin{cases} \dfrac{\log_e(1+x)}{x}, & x\neq 0,\\[4pt] 1,&x=0. \end{cases}$$

To decide whether the function increases or decreases we compute its first derivative. For the quotient $$\,f(x)=\dfrac{\log_e(1+x)}{x}\;(x\neq 0)$$ we use the Quotient Rule, which says $$\left(\dfrac{u}{v}\right)'=\dfrac{u'v-u\,v'}{v^{2}}.$$ Here $$u=\log_e(1+x),\;v=x.$$ Their derivatives are $$u'=\dfrac{1}{1+x},\;v'=1.$$ Applying the rule we obtain

$$f'(x)=\frac{\dfrac{1}{1+x}\,x-\log_e(1+x)\cdot 1}{x^{2}} =\frac{\displaystyle\frac{x}{1+x}-\log_e(1+x)}{x^{2}} \qquad(x\neq 0).$$

The denominator $$x^{2}$$ is always positive for every $$x\neq 0,$$ hence the sign of $$f'(x)$$ is the sign of the numerator

$$g(x)=\frac{x}{1+x}-\log_e(1+x),\qquad -1<x<\infty,\;x\neq 0.$$

Because $$1+x>0$$ on the whole domain, multiplying by the positive number $$1+x$$ keeps the sign unchanged. So we define

$$h(x)=(1+x)\,g(x)=x-(1+x)\log_e(1+x).$$

The signs of $$g(x)$$ and $$h(x)$$ are identical, and $$h(x)$$ is a little easier to study. First notice

$$h(0)=0-(1+0)\log_e(1+0)=0-0=0.$$

Next we differentiate $$h(x)$$:

$$h'(x)=1-\Bigl[\log_e(1+x)+(1+x)\cdot\dfrac{1}{1+x}\Bigr] =1-\log_e(1+x)-1 =-\log_e(1+x).$$

The natural logarithm behaves as follows:

• For $$x>0$$ we have $$\log_e(1+x)>0,$$ hence $$h'(x)=-\log_e(1+x)<0.$$ So $$h(x)$$ is decreasing on $$(0,\infty).$$
• For $$-1<x<0$$ we have $$\log_e(1+x)<0,$$ hence $$h'(x)=-\log_e(1+x)>0.$$ So $$h(x)$$ is increasing on $$(-1,0).$$

Because $$h(0)=0,$$ and $$h(x)$$ increases as we move left from $$0$$ (but remains negative, as we shall now see) while it decreases as we move right of $$0,$$ the maximum possible value of $$h(x)$$ is $$0.$$ Indeed, pick any $$x\in(-1,0):$$ since $$h$$ is increasing when we go towards $$0,$$ we must have $$h(x)<h(0)=0.$$ Similarly, pick any $$x>0:$$ because $$h$$ is decreasing away from $$0,$$ we again have $$h(x)<h(0)=0.$$ Therefore

$$h(x)\lt 0\qquad\text{for every }x\in(-1,\infty),\;x\neq 0.$$

As a consequence

$$g(x)=\frac{h(x)}{1+x}\lt 0\qquad\text{(since }1+x>0).$$

Finally, recalling $$f'(x)=\dfrac{g(x)}{x^{2}}$$ and that $$x^{2}>0,$$ we conclude

$$f'(x)\;=\;\dfrac{g(x)}{x^{2}}\;<\;0\quad\text{for every }x\in(-1,\infty),\;x\neq 0.$$

At $$x=0$$ the derivative obtained by the limit of the difference quotient is also $$0$$ from both sides, so there is no change in sign. Because the derivative is never positive anywhere in the domain, the function is strictly decreasing on the entire interval $$(-1,\infty).$$

Hence, the correct answer is Option D.