Question

Find the minimum value of $f\left(x\right)=\frac{e^x}{[x+1]},x\ge 0$,

Answer 1

Assuming that $x+1$ is integer we get

$f\left(x\right)=\frac{e^x}{[x+1]}$ $=\frac{e^x}{x+1}$

Now
$\frac{df\left(x\right)}{dx}=\frac{(x+1)e^x-e^x}{(x+1{)}^2}$ ..... Using $\frac{d\left(\frac{u}{v}\right)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
$=\frac{e^{x}x}{(x+1{)}^2}$

Now

$f^{\prime }\left(x\right)>0$ implies
$\frac{e^{x}x}{(x+1{)}^2}\ge 0$
Or
$x\ge 0$

Hence $f\left(x\right)$ is a strictly increasing function for all $xϵR^{+}$

The minimum value of the function will be at $x=0$.

Hence $f_{min}=f\left(0\right)$
$f\left(0\right)=\frac{e^0}{\left[1\right]}$
$=e^0$ $=1$

Thus the minimum value of the function is 1.