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

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Homework Help Overview

The problem involves finding a positive number that minimizes the expression x² + (1/x²). Participants are exploring calculus concepts related to optimization and critical points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • One participant presents a derivative approach to find critical points and suggests that the solution may be x=1. Others question whether this is the only solution and discuss the nature of the critical point in relation to minima.

Discussion Status

Participants are engaged in verifying the correctness of the solution and exploring the implications of the critical point found. There is a focus on confirming whether the identified point is indeed a minimum and if other values could potentially yield a smaller result.

Contextual Notes

There is an emphasis on the requirement for the solution to be a positive number, and some participants are considering the implications of the function's concavity in relation to the minimum value.

yoyo16
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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 :)
 
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[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.
 
Is it possible for the minimum to be a smaller value or would the smallest value be 1?
 
yoyo16 said:
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?
 
yoyo16 said:

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

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