Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 33

Given: f(x)=2x+(1/x)=2x+1x^-1
Find the critical values of x by setting the derivative of the function equal to zero and solving for the x-value(s).
f'(x)=2-1/x^2=0
2=1/x^2
2x^2=1
x^2=1/2
x=+-.707
The critical value is at x=.707, x=-.707 and x=0. The original function f(x) is not defined at 0.
If f'(x)>0, then the function is increasing on that interval.
If f'(x)<0, then the function is decreasing on that interval.
Choose any value for x that is less than -.707.
f'(-1)=1 Since f'(-1)>0, the function is increasing on the interval (-oo, -.707).

Choose any value for x that is in the interval (-.707,0).
f'(-.5)=-2 Since f'(-.5)<0, the function is decreasing on the interval (-.707,0).
Choose any value for x that is in the interval (0, .707).
f'(.5)=-2 Since f'(.5)<0, the function is decreasing on the interval (0, .707).
Choose any value for x that is greater that .707.
f'(1)=1 Since f'(1)>0, the function is increasing on the interval (.707, oo).
Since the sign of the derivative changed from positive to negative, there will be a relative maximum at x=-.707. The relative maximum is the point (-.707, -2.828).
Since the sign of the derivative changed from negative to positive, there will be a relative minimum at x=.707. The relative minimum is the point (.707, 2.828).