Maximum and minimum of a function existence

Question

Show $f: R_+→R$, $f(x) = \frac 1 {1+x}$ has no max or min.

The function is bounded since it is strictly decreasing on positive R and both limit exists when x=0 or $\infty$ but since R+ is an open set and 0 and $\infty$ is not part of the set, there is no max or min but only inf and sup. It seems really intuitive but how do i show that max or min does not exist more rigorously, possibly using definitions?

Answer 1

$f'(x)={{-1}\over{(x+1)^2}}$ is never $0$ we deduce that $f$ cannot have an extremum since the derivative is zero at an extremum.

Answer 2

Since $x>0$, you have $1+x>0$, and $\frac1{1+x}>0$.

And $1+x>1$, so $\frac1{1+x}<1$.

Equality is not achieved: if $\frac1{1+x}=1$, then $x=0$. And $\frac1{1+x}=0$ is impossible.

The existence of the limits, together with continuity, guarantees that you can find values as close to $0$ and $1$ as you want. So the range is $(0,1)$.