Domain of Square Root Function: Solving Inequalities for x ≤ -2

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The discussion focuses on finding the domain of the square root function defined by the inequality 1/(x+1) - 4/(x-2) ≥ 0. The initial steps involved cross-multiplying and simplifying to arrive at the inequality -3(x+2) ≥ 0, leading to the conclusion that x ≤ -2. However, participants clarify that the domain is not limited to this restriction alone, as the function's behavior must be considered across its entire range. It is suggested to multiply by the square of the denominator to accurately determine the domain. Ultimately, the conclusion is that the domain extends beyond just x ≤ -2.
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I am trying to find the domain of a square root function... To do so I have to solve the following inequality:


1/(x+1) - 4/(x-2) >= 0

This is how i attempted to solve it...:

I crossmultilplied the denominator to get

[(x-2) - 4(x+1)]/(x-2)(x+1) >= 0

Multiplied both sides by (x-2)(x+1)

(x-2) - 4(x+1) > = 0

Expanded

x - 2 -4x - 4 = 0

-3x -6 >= 0

-3(x+2) >= 0

(x+2) <= 0 <---- at this point I am not sure if i swap the sign around, I haven't been taught inequalities before... but I will swap it around anyway.

x <= -2

Is this the correct answer? When I graph the entire function (sqrt of the above), I get part of the function less than -2 but also part greater than -2... I don't really understand how there can be x > -2 if I got this restriction here of >-2.
 
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When dealing with an inequality, if you multiply by a negative number, the inequality changes.

You can deal with this by multiplying by the square of the denominator

i.e. ((x+1)(x-2))2
 
Yeah, when I calculate it I get the same answer:

x <= (-2)
 
Just as rock.freak667 said, multiply both sides by (x+1)2(x-2)2. The domain is not just x <= -2.
 

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