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AS C1 Inequalities

2/(x-3) < 1

(that's supposed to be a 'less than or equal to sign', not a 'less than' sign)

My natural reaction is to multiply everything by (x-3) and work from there, but apparently I'm supposed to multiply everything by (x-3)^2. Can anyone explain why?
Reply 1
Avatar for henryt
henryt
17
Because if the expression (x-3) is negative, you would have to change the sign of the inequality. However, squaring it obviously gives you an answer which is always positive, so no ambiguity!! Hurrah!
Reply 2
Avatar for Knogle
Knogle
16
(x-3) may have a negative value, in which case when you multiply it accross the equation, the direction of the sign should change to >.

In order to ensure that you're multiplying by a 0/+ve value, you use (x-3)2.
Aha, that's smart. Thanks people.
Reply 4
Avatar for EDerkatch
EDerkatch
Yes but squaring a function can sometimes create extra roots

2/(x-3)-1<0

(-x+5)/(x-3)<0

The above function has a root at x=5 and an asymptote at x=3 and is negative for values less than 3 and greater than 5

x<3 and x>5
Reply 5
Avatar for Laws
Laws
why not just multiply by (x-3) and flip the inequality sign around?
Reply 6
Avatar for Knogle
Knogle
16
Laws
why not just multiply by (x-3) and flip the inequality sign around?

Because (x-3) may be positive, in which case you shouldn't flip the inequality sign around.
The Ace is Back
2/(x-3) < 1

(that's supposed to be a 'less than or equal to sign', not a 'less than' sign)

My natural reaction is to multiply everything by (x-3) and work from there, but apparently I'm supposed to multiply everything by (x-3)^2. Can anyone explain why?

Here's a good method.

2/(x-3) - 1 < 0
(2-(x-3))/(x-3) < 0
(5-x)/(x-3) < 0

Let f(x) = (5-x)/(x-3)
and in the intervals x<5, 5<x<3, x>3 check the sign of f(x)
That avoids the sign of the denominator problem.
The Ace is Back
2/(x-3) < 1

(that's supposed to be a 'less than or equal to sign', not a 'less than' sign)

My natural reaction is to multiply everything by (x-3) and work from there, but apparently I'm supposed to multiply everything by (x-3)^2. Can anyone explain why?

Here's a good method.

2/(x-3) - 1 < 0
(2-(x-3))/(x-3) < 0
(5-x)/(x-3) < 0

Let f(x) = (5-x)/(x-3)
and in the intervals x<3, 3<x<5, x>5 check the sign of f(x)
That avoids the sign of the denominator problem.

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