How Do You Solve the Inequality \(\frac{1}{2^x} > \frac{1}{x^2}\)?

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Homework Help Overview

The discussion revolves around solving the inequality \(\frac{1}{2^x} > \frac{1}{x^2}\), which can be transformed into the equivalent form \(x^2 > 2^x\). Participants are exploring the behavior of the functions involved and their intersections.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to identify points of intersection, noting that they know one intersection occurs at \(x = 2\). Others express uncertainty about finding additional intersection points, referred to as \(a\) and \(b\), and discuss the implications of these points on the solution regions.

Discussion Status

The conversation is active, with participants sharing insights on how to approach the problem. Some suggest checking specific regions to determine where the inequality holds, while others acknowledge the difficulty in finding certain intersection points explicitly. There is recognition that graphical methods may provide insight into the locations of intersections.

Contextual Notes

Participants mention the challenge of finding negative intersections and the potential use of the Lambert W function for approximations, indicating a level of complexity in the problem that may not be solvable with elementary functions alone.

zeion
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Homework Statement



Use your knowledge of exponents to solve

[tex]\frac{1}{2^x} > \frac{1}{x^2}[/tex]


Homework Equations





The Attempt at a Solution



[tex]x^2 > 2^x[/tex]

Then I am stuck.

I know they intersect at x = 2.
 
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Try to find out when the equality holds. Let's say they hold at a and b. This will give you regions [tex]]-\infty,a[, ]a,b[, ]b,+\infty[[/tex]. From any region, pick a point and check if the inequality is satisfied at that point. If so, then that region is part of the solution.
 
micromass said:
Try to find out when the equality holds. Let's say they hold at a and b. This will give you regions [tex]]-\infty,a[, ]a,b[, ]b,+\infty[[/tex]. From any region, pick a point and check if the inequality is satisfied at that point. If so, then that region is part of the solution.

Thing is I don't know how to find what a and b are. (Except for 2)
 
Well, yeah, I don't think you can explicitly find the value of a and b (except for 2). But you know from the graph that such an a and b exist and where they lie.
 
zeion said:

The Attempt at a Solution



[tex]x^2 > 2^x[/tex]

Then I am stuck.

I know they intersect at x = 2.

They intersect at x=2 because [itex]2^2 = 2^2[/tex]<br /> <br /> What about x=4?<br /> Does [itex]4^2 = 2^4[/tex] ?[/itex][/itex]
 
Yes, but how do you find the negative intersection?
 
You approximate. And if you're in a higher college math class, you find it in terms of the Lambert W function. Either way, you can't express the root of x2=2x, x<0 explicitly in terms of elementary functions.
 

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