How do you find the critical points for # x+2x^-1# and the local max and min?

1 Answer
Apr 7, 2018

Answer:

Critical pointss are at #(-sqrt2,-2sqrt2)# and #(sqrt2,2sqrt2)#. At #(-sqrt2,-2sqrt2)# we have a local maxima and at #(sqrt2,2sqrt2)# we have a local minima.

Explanation:

Critical points for a function #y=f(x)# appear where #(dy)/(dx)=0#. At these points we have local maxima and minima too. The maxima is when #(d^2y)/(dx^2)<0# and minima is when #(d^2y)/(dx^2)>0#.

As #y=x+2x^(-1)#

#(dy)/(dx)=1-2/x^2# and #(d^2y)/(dx^2)=6/x^3#

#(dy)/(dx)# is zero when #1-2/x^2=0#

or #x^2-2=0# i.e. #x=+-sqrt2#, hence we have critical points at #x=-sqrt2# (where #f(x)=-sqrt2-2/sqrt2=-2sqrt2#) and at #x=sqrt2# (where #f(x)=sqrt2+2/sqrt2=2sqrt2#) and hence coordinates are #(-sqrt2,-2sqrt2)# and #(sqrt2,2sqrt2)#.

At #x=-sqrt2#, #(d^2y)/(dx^2)=-6/(2sqrt2)<0# hence we have a local maxima

and at #x=sqrt2#, #(d^2y)/(dx^2)=6/(2sqrt2)>0# hence we have a local minima.

graph{x+2x^(-1) [-10, 10, -5, 5]}