
#1
Mar1410, 12:24 AM

P: 122

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/(x2) >= 0 This is how i attempted to solve it...: I crossmultilplied the denominator to get [(x2)  4(x+1)]/(x2)(x+1) >= 0 Multiplied both sides by (x2)(x+1) (x2)  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 dont really understand how there can be x > 2 if I got this restriction here of >2. 



#2
Mar1410, 12:36 AM

HW Helper
P: 6,214

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)(x2))^{2} 



#3
Mar1710, 01:39 AM

P: 61

Yeah, when I calculate it I get the same answer:
x <= (2) 



#4
Mar1710, 09:39 AM

Mentor
P: 21,062

Solving an inequality
Just as rock.freak667 said, multiply both sides by (x+1)^{2}(x2)^{2}. The domain is not just x <= 2.



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