Question

The maximum value of the function $f\left(x\right)=ln(1+x)-x$ (where $x>-1$) occurs at $x=$

• A. 0

Answer 1

The correct option is A 0
$f\left(x\right)=ln(1+x)-x,x>-1$
$f^{\prime }\left(x\right)=\frac{1}{1+x}-1=-\frac{x}{1+x}$
$f^{\prime }\left(x\right)=0$
$\Rightarrow -\frac{x}{1+x}=0$
$x=0$
$f^{\prime \prime }\left(x\right)=-\left[\frac{(1+x)\left(1\right)-x\left(1\right)}{(1+x{)}^2}\right]$
$f^{\prime \prime }\left(0\right)=-1<0$
$\Rightarrow f\left(x\right)$ has maxima at $x=0$