Question

For $x\ge 0,$ the minimum value of $f\left(x\right)=\ln (1+x)-x+\frac{x^2}{2}$ is

Answer 1

Given : $f\left(x\right)=\ln (1+x)-x+\frac{x^2}{2}$
Differentiating the above w.r.t. $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x}-1+x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2}{1+x}\ge 0\forall x\ge 0$
$\Rightarrow f\left(x\right)$ is an increasing function $\forall x\ge 0$
And
$f\left(0\right)=\ln (1+0)-0+\frac{0^2}{2}=0$
Since $x\ge 0\Rightarrow f\left(x\right)\ge f\left(0\right)$
$\Rightarrow f\left(x\right)\ge 0$
So, the minimum value of $f\left(x\right)$ is $0.$