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

1 Answer
Shwetank Mauria
Apr 7, 2018

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]}

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