Does $1 + \frac{1}{x} + \frac{1}{x^2}$ have a global minimum for $0 < x \in \mathbb{R}$?

Question

This question is related to this one.

My question here is:

Does the function

$$f(x) = 1 + \frac{1}{x} + \frac{1}{x^2}$$

have a global minimum for $0 < x \in \mathbb{R}$?

Thank you!

Answer 1

Completing the square, $f(x)=\frac{3}{4}+(\frac{1}{x}+\frac{1}{2})^2$, so $x=-2$ is a global minimum, and there is no minimum, absolute or relative, for $x\gt -1$.

Answer 2

no. it's obviously bigger than 1 for any positive x, but it's limit as $x \rightarrow \infty$ is 1, so your expression can be as close to 1 as you wish, but it never attains 1 for any x

Answer 3

The first derivative is $ \ f'(x) \ = \ - \ x^{-2} \ - \ 2 x^{-3} \ , $ which is only zero for $ \ x = -2 \ $ ; the other critical point is at $ \ x = 0 \ , $ where the function is undefined. So there is no relative minimum for positive values of $ \ x \ . $ And, as already pointed out by mm-aops , the function only approaches a "horizontal asymptote" for $ \ x > 0 \ , $ so there is no global minimum to be found.