Range of validity for Binomial Series

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The discussion focuses on finding the range of validity for two binomial series: (1 + 3x/2)^(-1) and (1 + 1/(3x))^(-1). For the first series, the valid range is determined to be -2/3 < x < 2/3. The second series requires careful consideration of the sign of 3x when solving inequalities, leading to two cases: x > 1/3 when 3x is positive and x < -1/3 when 3x is negative. The importance of correctly handling the direction of inequalities based on the sign of the variable is emphasized. The discussion concludes with a clearer understanding of the range for the second series.
LiHJ
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Homework Statement



Dear Mentors and Helpers,
here's the question:
Find the range of validity for (1 + 3x/2)^(-1) and (1 + 1/(3x))^(-1).

Homework Equations

The Attempt at a Solution


For the first binomial series:

-1 < 3x/2 < 1
-2 < 3x < 2 (multiply 2 throughout)
-2/3 < x < 2/3 (divide by 3 throughout)

For the second binomial series:
-1 < 1/(3x) < 1
-3x < 1 < 3x (multiply 3x throughout)

-3x < 1 or 1 < 3x
x > -1/3 (divide -3, inequality change sign) or 1/3 < x (divdide 3)

Therefore I get: x > -1/3 or x > 1/3

However my range isn't correct for the second binomial series can any Mentors or PF helper guide me and correct my mistakes.

Thank you
 
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For the second series you multiply each part of the inequality by 3x. When you multiply by -3, you note that it is negative so you change the direction of the inequality- but 3x might be negative also. So you must also consider the sign of 3x.

If 0< 1/3x< 1 then 3x is positive so 0< 1< 3x, x> 1/3. If -1< 1/3x< 0 then 3x is negative so -3x> 1> 0 and x<-1/3, not ">".
 
Thank you for the explanation, I finally understand :w
 

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