inequality (1 Viewer)

[untouchablecuz]

I never said that I was (Interestingly enough though, I can still do simple inequalities )

you're a wanker

 

[cutemouse]

you're a wanker

Don't talk to yourself --> It's a bad habit.

 

[frenzal_dude]

1/(3x+1)>1/(2x-3)

(2x-3)^2(3x+1)>(3x+1)^2(2x-3)

(2x-3)^2(3x+1)-(3x+1)^2(2x-3)>0

(2x-3)(3x+1)(2x-3-3x-1)>0

-(2x+3)(3x+1)(x+4)>0

by graphing or testing value we get x<-4, -1/3 < x < 3/2

So we know that (2x+3)(3x+1)(x+4)<0 = so either:
1. all 3 are -ve , or
2. only 2x+3 < 0, or
3. only 3x+1 < 0, or
4. only x+4 < 0, then we get:

1. x<-4
2. x<-1/3 and x>-4
3. x>-1/3 and x<3/2
4. x>-1/3 and x<-4

How do you work out without graphing that 1. and 3. are the correct answers?

 

[cutemouse]

So we know that (2x+3)(3x+1)(x+4)<0 = so either:
1. all 3 are -ve , or
2. only 2x+3 < 0, or
3. only 3x+1 < 0, or
4. only x+4 < 0, then we get:

1. x<-4
2. x<-1/3 and x>-4
3. x>-1/3 and x<3/2
4. x>-1/3 and x<-4

How do you work out without graphing that 1. and 3. are the correct answers?

That's a very annoying way of doing this (especially because you now have 3 factored terms). So you would need to take 6 cases, not just the 4 that you've mentioned I believe.

 

[frenzal_dude]

That's a very annoying way of doing this (especially because you now have 3 factored terms). So you would need to take 6 cases, not just the 4 that you've mentioned I believe.

Hi, so what would be the 4th and 5th case?
for it to be <0 all 3 should be -ve coz 3 -ve numbers multiplied will give -ve, or if just one of each are negative and the other 2 +ve, you'll still get <0, so shouldn't there only be 4 cases?

How would you work this out without graphing or testing values?

 

[cutemouse]

Hi, so what would be the 4th and 5th case?
for it to be <0 all 3 should be -ve coz 3 -ve numbers multiplied will give -ve, or if just one of each are negative and the other 2 +ve, you'll still get <0, so shouldn't there only be 4 cases?

I'm not too sure since I've never really used this method. But I was under the impression that you need 6 cases. (ie. all possible combinations of negative and positive cases)

How would you work this out without graphing or testing values?

Taking cases, like you've done. But why would you? It's like me saying that I'll always use the first principle approach of differential calculus to find the derivative of a function (instead of using the result).

 

[hscishard]

Man, get over it. Many people couldn't solve it. Eg: today i momentarily forgot how to do implicit differentiation in my math test but it came back to me. Monsterman may have experienced the same thing but just for a longer time and under the stress of doing well in the first quiz he probs forgot some stuff. It's natural, doesn't mean his noob or anything.

Is implicit differentiation is part of the 3 unit syllabus? It's under a heading: -Extension: Implicit differentiation- in the cambridge book.

 

[mirakon]

It's technically 4 unit.

 

[shaon0]

Is implicit differentiation is part of the 3 unit syllabus? It's under a heading: -Extension: Implicit differentiation- in the cambridge book.

Best to learn it. It's not very hard and very useful for uni as they won't teach you it during initial coursework.

 

[cutemouse]

Is implicit differentiation is part of the 3 unit syllabus? It's under a heading: -Extension: Implicit differentiation- in the cambridge book.

Nope. But it's not hard. Probs wouldn't hurt learning it.

 

[dasicmankev]

1/(3x+1) > 1/(2x-3)

1/(3x+1) - 1/(2x-3) > 0

Using common denominator:

(-x-4)/[(3x+1)(2x-3)] > 0

Multiply inequality on both sides by denominator squared:

-(x+4)(3x+1)(2x-3) > 0

Sketch curve noting the branches above x-axis. This will give the answer of x<-4, -1/3 < x < 3/2