Question

Let x be real, x>0 and set f(x)=x+1/x. Find the absolute min for f(x).

Answer 1

f'(x) = 0)
1 - 1/x^2 = 0 => x^2 - 1 = 0
since x>0, x=1
you have minimum at this value because either you take x->0 or x->inf
f(x) -> inf, and there are no other stationary points

Answer 2

f''(1) = 2 > 0 ... you can say it min from second derivative test.