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))#
