For what value of x does (1/x) + (1/(1-x)) > 0

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The discussion centers on solving the inequality (1/x) + (1/(1-x)) > 0. It is established that x cannot equal 0 or 1, and the expression is positive when 0 < x < 1. The user finds that the solution manual incorrectly includes x > 1 as a valid range. Testing values confirms that the inequality does not hold for x > 1, reinforcing confidence in the user's solution. The conclusion is that the solution manual contains a printing error, affirming the correct range is 0 < x < 1.
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Homework Statement



For what value of x does \frac{1}{x} + \frac{1}{1-x} &gt; 0

Homework Equations


The Attempt at a Solution



So I see that x not = 0 and x not = 1.
Then I added the fractions and see that it will only be positive if the bottom is positive
ie. x(1-x) > 0
I define f(x) = x(1-x) that is a parabola.
It will change signs at x = 0 and x = 1.
So I test some values in the intervals and see that it is positive when 0 < x < 1
So that is the answer.
But the solution says that it is 0 < x < 1 or x > 1?
If I put in 2 > 1 I will get (1/2) + (1/(1-2)) = (1/2)+(1/(-1) = -1/2 not > 0
 
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Solution manuals do make mistakes. It sounds like you have a good handle on this proof technique, and have taken advantage of some sanity checks (another would be to graph 1/x + 1/(1-x)) that support your answer and conflict with the manuals, so you would have good reason to remain confident in your solution.
 
There's nothing to worry about,
Your answer is correct, there must have been some printing error in the solution.
 
It seems that the solution manual is incorrect.

edit: began my original response before others posted. Carry on...
 
Zeion pwned the textbook.
 
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