the derivative dilemma

Let $f(x) = x + \frac{1}{x}$ .

Thus, $f'(x) = 1 - \frac{1}{x^{2}}$ .

Solving $f'(x) = 0$ gives us $x = \pm 1$ , which gives us the critical points of the function.

Taking the second derivative of $f(x)$ yields $f''(x) = -\frac{2}{x^{3}}$ ; for $x = 1$ , we have $f''(x) = -2$ ; since it is negative, $f(1)$ is a local minimum and we have the minimum value of $f(x)$ at $x = 1$ , which is $2$ .

Using the A.M.-G.M. inequality, we have

$\frac {x+\frac {1}{x}}{2} ≥ \sqrt{(x)\left(\frac {1}{x} \right)}$

$\boxed {x+\frac {1}{x} ≥ 2}$

When we consider x=0 then $\frac{1}{x}$ will become undefined so we must consider x=1 $=>x+\frac{1}{x}=1+\frac{1}{1}=2$ thatsolve