$$f(x)=x+\frac{1}{x}\\ \therefore f^{'}(x)=1-\frac{1}{x^2}\\ \text{now, } f^{'}(x)=0\ \Rightarrow x=+1,-1\\ f^{"}(x)=\frac{2}{x^3}\\f^{"}(1)>0\ \&\ f^{"}(-1)<0\\ \max f(x)=-2<\min f(x)=2$$ So, the funtion has maxima and minima respectively at $x=-1$ and $x=1$
But, function has far more value than the maxima and far less value than the minima. Moreover, minima is larger than maxima. I am confused. Thanking You.
The procedure of finding stationary point and verifying the second gradient only promises local optimality.
As we can see from the graph, $(1,2)$ indeed is a local minimum and $(-1,-2)$ is indeed a local maxima.
They are only smallest or largest in a confined region.
The shortest person in a group can be taller than the tallest person in another group.