Calculus - How to minimize x^2 + (1/x^2)

[yoyo16]

Homework Statement

Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

Homework Equations

The Attempt at a Solution

y=X^2+(1/X^2)
Derivative=2x-2x^(-3)
0=2x-2x^3
2/x^3=2x
2=2x^4
1=x^4
x=1

The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)

 

[vela]

[strike]Looks good except that you missed a possible solution. What other value of x satisfies x^4=1?[/strike]

Oops, never mind. I didn't notice you were looking for a positive number.

You just now need to verify that the critical point corresponds to a minimum.

 

[yoyo16]

Is it possible for the minimum to be a smaller value or would the smallest value be 1?

 

[LCKurtz]

Is it possible for the minimum to be a smaller value or would the smallest value be 1?

If the graph is concave up for $x>0$ your relative minimum will be an absolute minimum on that interval. Have you checked that?

 

[Ray Vickson]

Homework Statement

Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

Homework Equations

The Attempt at a Solution

y=X^2+(1/X^2)
Derivative=2x-2x^(-3)
0=2x-2x^3
2/x^3=2x
2=2x^4
1=x^4
x=1

The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)

Your function is strictly convex in the convex region {x > 0}, so any stationary point is automatically a global minimum within the region.

 

[yoyo16]

Oh ok so since it is absolute it would be 1 right?