VCE Stuff > VCE Specialist Mathematics
pi's Specialist Maths Questions
pi:
--- Quote from: TrueTears on July 25, 2011, 07:15:15 pm ---
--- Quote from: Rohitpi on July 25, 2011, 07:07:42 pm ---Is it due to the fact that x {-1, 0} from the denominator, and hence x can exist between these asymptotes?
I'm not really sure tbh :(
--- End quote ---
Good guess, but it's something simpler!
Think about what you are doing precisely when you "cross-multiplied" to arrive at 4x =>x+1
First you would compute 2(2x) and then you would compute 1(x+1), but who says (2x) is positive and (x+1) is positive? Thus we can not cross multiply in this case since it could change the sign of the inequality!
--- End quote ---
Oh. So does that mean I'll have to solve with the equations you outlined earlier instead:
--- Quote from: TrueTears on July 25, 2011, 03:15:06 pm ---Now just split into cases, either and or and
--- End quote ---
Simultaneously? Or using quick sketch-graphs for the quadratics and finding the intersections of the values for each pair?
I'm still confused on how the inequality sign changed. :(
TrueTears:
--- Quote from: Rohitpi on July 25, 2011, 07:25:42 pm ---
--- Quote from: TrueTears on July 25, 2011, 07:15:15 pm ---
--- Quote from: Rohitpi on July 25, 2011, 07:07:42 pm ---Is it due to the fact that x {-1, 0} from the denominator, and hence x can exist between these asymptotes?
I'm not really sure tbh :(
--- End quote ---
Good guess, but it's something simpler!
Think about what you are doing precisely when you "cross-multiplied" to arrive at 4x =>x+1
First you would compute 2(2x) and then you would compute 1(x+1), but who says (2x) is positive and (x+1) is positive? Thus we can not cross multiply in this case since it could change the sign of the inequality!
--- End quote ---
Oh. So does that mean I'll have to solve with the equations you outlined earlier instead:
--- Quote from: TrueTears on July 25, 2011, 03:15:06 pm ---Now just split into cases, either and or and
--- End quote ---
Simultaneously? Or using quick sketch-graphs for the quadratics and finding the intersections of the values for each pair?
I'm still confused on how the inequality sign changed. :(
--- End quote ---
U mean how did the inequality sign change in my working? It didn't change, i just split it into cases. It's the exact same concept as solving quadratic inequalities, but here we are solving rational function inequalities, to see how to solve them check here http://vce.atarnotes.com/agora/index.php/topic,39795.msg416590.html#msg416590
With regards to your question, yeah you have to solve my inequalities to yield the correct answer, but besides knowing what inequalities to solve, the more important thing is to realise why you can't solve it the way outlined earlier.
pi:
What I meant was the sign change in -1<x<0, in the equations you have (and I had, although I did it with the positive assumption, which I see is clearly wrong) they were . I'm not sure on the algebraic way to get this. I got it graphically (kinda), but algebraically, I'm just not sure.
OH, I GET IT NOW! GREAT LINK :) :) :)
(i think I had them on my desktop though, next time I'll d/l + READ ;) )
pi:
Quick q, hyperbolic trig functions (and their inverses) are definitely NOT on the course right?
(our SACs think otherwise :( , gotta ask)
b^3:
As far as I know, no they are not, and we haven't covered them either.
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