Question

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

• A. increases in $(-1,\infty )$
• B. decreases in $(-1,0)$ and increases in $(0,\infty )$
• C. increases in $(-1,0)$ and decreases in $(0,\infty )$
• D. decreases in $(-1,\infty )$.

Answer 1

The correct option is D decreases in $(-1,\infty )$.

$f\left(x\right)=\frac{1}{x}\ln (1+x)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1-\frac{1}{1+x}-\ln (1+x)}{x^2}$
$f^{\prime }\left(x\right)<0\forall x\in (-1,\infty )$