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

$f(x) = {1 \over x}{\log _e}\left( {1 + x} \right)$<br><br> $$ \Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}$$<br><br> $$ \Rightarrow f'(x) = {{x - \left( {1 + x} \right){{\log }_e}\left( {1 + x} \right)} \over {{x^2}\left( {1 + x} \right)}}$$<br><br> Let $g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)$<br><br> $$ \Rightarrow g'(x) = 1 - \left( {1 + x} \right){1 \over {1 + x}} - \left( 1 \right) \times {\log _e}\left( {1 + x} \right)$$<br><br> $= 1 - 1 - {\log _e}\left( {1 + x} \right)$<br><br> $= - {\log _e}\left( {1 + x} \right)$<br><br> For $x \in \left( { - 1,0} \right),g'(x) &gt; 0$<br> <br>and for $x \in \left( {0,\infty } \right),g'(x) &lt; 0$<br><br> Also, $g(0) = 0 - \left( {1 + 0} \right){\log _e}\left( {1 + 0} \right) = 0$<br><br> $\therefore g'(x) &lt; 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$<br><br> $\Rightarrow f'(x) &lt; 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$<br><br> $\Rightarrow$ f(x) is a decreasing function for all $x \in \left( { - 1,\infty } \right)$