Let f(x)=y
y=2x+(5/x)
dy/dx=2-(5/x^2)
At the local extremas dy/dx=0
2-5/x^2=0
Solve for x
2-5/x^2=0|*x^2
2x^2-5=0|+5|:2
x^2=5/2|sqrt()
x=+-sqrt(5/2)=+-sqrt(10)/2
x_1=sqrt(10)/2 or x_2=-sqrt(10)/2
Testing to see which one is minimum and maximum we find the second derivative. If (d^2y)/(dx^2)>0, it is a local minimum. If (d^2y)/(dx^2)<0, it is a local maximum.
(d^2y)/(dx^2)=10/x^3
x_2=-sqrt(10)/2
(d^2y)/(dx^2)=-10/(sqrt(10)/2)^3<0
Since it is less than 0 the local maxima is at x_2=-sqrt(10)/2
When x=-sqrt(10)/2,y=-2sqrt(10).
Therefore local maxima is at p_(max)(-sqrt(10)/2|-2sqrt(10))
When x=sqrt(10)/2,(d^2y)/(dx^2)=(4/5)sqrt(10)>0. Hence local minimum is at x=sqrt(10)/2
When x=sqrt(10)/2, y=2sqrt(10)
Hence local minimum is at p_(min)(sqrt(10)/2|2sqrt(10))
