More Curve Sketching (1 Viewer)

[cutemouse]

I'm pulling hairs with this one!!! I cannot seem to factorise the second derivative and cannot find the point of inflexion. There's meant to be one at x=0, but...?!?!?!

Anyway, the question is:

Sketch, showing any stationary points, points of inflexion or asymptotes:

y=(2x+3)/(x²-4)

Cheers

 

[Trebla]

I'm pulling hairs with this one!!! I cannot seem to factorise the second derivative and cannot find the point of inflexion. There's meant to be one at x=0, but...?!?!?!

Anyway, the question is:

Sketch, showing any stationary points, points of inflexion or asymptotes:

y=(2x+3)/(x²-4)

Cheers

y = (2x + 3) / (x² - 4)
u = 2x + 3, u' = 2
v = x² - 4, v' = 2x
dy/dx = [2(x² - 4) - 2x(2x + 3)] / (x² - 4)²
= [2x² - 8 - 4x² - 6x] / (x² - 4)²
= (- 2x² - 6x - 8) / (x² - 4)²
= - 2(x² + 3x + 4) / (x² - 4)²
u = x² + 3x + 4, u' = 2x + 3
v = (x² - 4)², v' = 4x(x² - 4)
d²y/dx² = - 2 [(2x + 3)(x² - 4)² - 4x(x² - 4)(x² + 3x + 4)] / (x² - 4)⁴
Inflexion point occurs when d²y/dx² = 0
i.e. (2x + 3)(x² - 4)² - 4x(x² - 4)(x² + 3x + 4) = 0
(x² - 4)[(2x + 3)(x² - 4) - 4x(x² + 3x + 4)] = 0
(x² - 4)[(2x + 3)(x² - 4) - 4x(x² + 3x + 4)] = 0
(x² - 4)[2x³ - 8x + 3x² - 12 - 4x³ - 12x² - 16x)] = 0
(x² - 4)(- 2x³ - 9x² - 24x - 12) = 0
(x² - 4)(2x³ + 9x² + 24x + 12) = 0
x² - 4 = 0 yields x = ± 2 but the function is undefined for x = ±2
2x³ + 9x² + 24x + 12 = 0 yields an ugly number for x....lol

I've computed a sketch of the graph and can confirm the inflexion point doesn't occur at x = 0, so you may have typed the question wrong?

 

[cutemouse]

Thank you for your help Trebla... But how do you know that the pt of inflexion is at x=0? Which rule states this?

 

[lyounamu]

Thank you for your help Trebla... But how do you know that the pt of inflexion is at x=0? Which rule states this?

Point of inflexion is located where the second derivative is equal to 0.

 

[cutemouse]

Ahh yes, now I get it. I forgot that x=0 was a result.

Thanks trebla. Lifesaver man!

 

[cutemouse]

y = (2x + 3) / (x² - 4)
dy/dx = [2(x² - 4) - 2x(2x + 3)] / (x² - 4)²
= [2x² - 8 - 4x² - 64] / (x² - 4)²
= (- 2x² - 56) / (x² - 4)²

Hey mate I think you made a mistake there... It think it should be -6x instead of -64..

I'm still pulling hairs on this one lol

 

[lyounamu]

Hey mate I think you made a mistake there... It think it should be -6x instead of -64..

I'm still pulling hairs on this one lol

Yeah. Now you got it, keep at it.

 

[cutemouse]

lol, I can't figure it out.

I get the 2nd derivative but you can't factorise it by x, therefore x=0 is not a factor. Yet there is a pt of inflexion at x=0...

Unless of course if you're allowed to times both sides by x when y''=0, but that'd be cheating

 

[lyounamu]

lol, I can't figure it out.

I get the 2nd derivative but you can't factorise it by x, therefore x=0 is not a factor. Yet there is a pt of inflexion at x=0...

Unless of course if you're allowed to times both sides by x when y''=0, but that'd be cheating

okay, i will have a go then..

If you diff twice, you get:

(8(2x+3)x^2)/(x^2-4)^3 - 8x/(x^2-4)^2 - (2(2x+3))/(x^2-4)^2

Then you let that equal to 0

so (8(2x+3)x^2)/(x^2-4)^3 - 8x/(x^2-4)^2 - (2(2x+3))/(x^2-4)^2 = 0
i.e. (8(2x+3)x^2 - 8x(x^2-4) - (2(2x+3))(x^2-4) = 0
simplify:
(16x+24)x^2 - 8x^3 + 32x - (4x+6)(x^2-4) = 0
16x^3 + 24x^2 - 8x^3 + 32x - (4x^3 + 6x^2 -16x - 24) = 0
4x^3 + 18x^2 + 48x + 24 = 0
2x^3 + 9x^2 + 24x + 12 = 0

USING newton's method, x is about -0.4255 where inflexion point is.

 

[cutemouse]

Eh? I've lost you...

I get y''=(4x³+18x²+48x+24)/(x²-4)³

Obviously x=0 isn't a factor. So...?

 

[lyounamu]

Eh? I've lost you...

I get y''=(4x³+18x²+48x+24)/(x²-4)³

Obviously x=0 isn't a factor. So...?

I am 100% sure my derivative is correct. I used one of the differentier to lessen the workload.

 

[cutemouse]

Yeah... and how exactly are you meant to get the solution x=0?

Hair pulling is not fun, esp. cos this is 2U stuff... hahaha

 

[Trebla]

I fixed my working up...got the same equation as lyounamu. Unless you got the question/answer wrong, I don't think this is 2 unit.

 

[cutemouse]

Yep, Maths in Focus has done it again. Their answers are wrong, like for nearly every single curve sketching question. The question however is correct.

Funny thing is, that in their older version they printed the correct answers and in the newer ones they printed the wrong ones...

 

[lyounamu]

Yep, Maths in Focus has done it again. Their answers are wrong, like for nearly every single curve sketching question.

Funny thing is, that in their older version they printed the correct answers and in the newer ones they printed the wrong ones...

That's why you never ever use maths in focus except for the challenge questions that they give occassionally.

I heard that Maths in focus is very good for explaining concepts but many of them are, in my opinion, too elementary. Questions are somewhat poor and little bit lower in terms of its standards in regards to HSC.

 

[cutemouse]

Yeah that's why I have a stack of Maths books that are as tall as Mt Everest, lol.

But if I can't get Math in Focus' questions, then what chance do I have for other harder questions?

I reckon they're absolutely crap for explaining theory. Coroneos FTW!

 

[lyounamu]

Yeah that's why I have a stack of Maths books that are as tall as Mt Everest, lol.

But if I can't get Math in Focus' questions, then what chance do I have for other harder questions?

I reckon they're absolutely crap for explaining theory. Coroneos FTW!

That doesn't mean much anyway. Some questions from MIF are poorly worded too hence appear to be harder or confusing in general. I might be generalising a bit here but from my own experience with that that's how I felt.

 

[cutemouse]

What book do you use for 2U questions?

 

[lyounamu]

What book do you use for 2U questions?

Fitz & camb & past papers (3 pillars of 2 unit maths)

 

[Trebla]

Jones & Couchman - I reckon the exercises are good practice
Cambridge Study Guide - Mainly cos my school used CSSA papers and it's gives a good indication of the standard of CSSA papers (as they're from the same people who write the CSSA trials)

However, I think textbooks can only offer so much. They're usually good for developing foundational skills as you learn a concept for the first time. Once you're familiar with the theory, textbook questions become really tedious...past exam questions are the way to go at that stage.