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

[Shaybay92]

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.

 

[rock.freak667]

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

 

[BL4CKCR4Y0NS]

Yeah, when I calculate it I get the same answer:

x <= (-2)

 

[Mark44]

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