Q14 (1 Viewer)

[sghguos]

What was the answer to these questions. Q14iii) Q14iv)

f(x) = 3x^4 + 4x^3-12x^2

Q14a) iii)

For what values of x was the function increasing.

My answer: left it blank can someone tell me how to do it


Q14a) iv) For what values of k will 3x^4 + 4x^3 - 12x^2 + k = 0

My answer: I had k<4

 

[rsagar]

Re: What was the answer to these questions. Q14iii) Q14iv)

I went dy/dx > 0 and worked it out from there...

Even i think I'm wrong... but i put something...

and I left Q14 iv) blank since i had no clue and it was only a one marker

 

[RealiseNothing]

Re: What was the answer to these questions. Q14iii) Q14iv)

14(a)iii) Just look at your graph.

iv) k > 32 iirc.

 

[Currybear]

Re: What was the answer to these questions. Q14iii) Q14iv)

it was >32

you had to draw the graph, and realise that you had a global min at (0,32)

 

[Demise]

Re: What was the answer to these questions. Q14iii) Q14iv)

x was increasing for -2<x<0 and >1.

 

[sghguos]

Re: What was the answer to these questions. Q14iii) Q14iv)

can anyone show me a picture of what the graph looked like?

I think I sketched it wrong oh god....

 

[RishBonjour]

http://www.wolframalpha.com/input/?i=3x^4+++4x^3+-12x^2

^ looks like that

 

[Timske]

Re: What was the answer to these questions. Q14iii) Q14iv)

can anyone show me a picture of what the graph looked like?

I think I sketched it wrong oh god....

well it should have looked like a concave upwards 'w'

 

[crazy_paki123]

can all you people who are saying that k>32 please answer this past hsc question then?
Q6 (C) iii, all working out
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/mathemat_01.pdf
your welcome

 

[Kimyia]

Put the equation from yesterday's paper into wolfram alpha or another graphing program. The lowest point is y= -32.
Then add +32 into the equation in place of k and graph again. Then the lowest point will be y=0 so if k>32, the curve doesn't cross the x-axis, hence no solutions when k>32.

 

[Kimyia]

I think to explain that 2001 question, the equation was made equal to k unlike yesterday's equation where they added a k. So if you did that with yesterday's equation, y would be equal to -k. So yes, there would be no solutions for y<-32 but then y = -k so -k<-32. Divide by a negative and change the sign gives you k>32.

 

[crazy_paki123]

yes but can you answer the 2001 question it is extremely similar, only that it is cubic. And the answer btw is B<=k<=A because that is the only region where the graph would intersect with a horizontal line 3 times. plus heres a link to confirm that
http://www.pasthsc.com.au/HSC_Mathematics_files/Maths2U01QuickAnswers.pdf
face it you guys lost a simple mark, just too ignorant to admit it

 

[megaman64]

for 14.a.ii. the graph sketching question. Did you have to write the x intercepts when drawing the graph? if you didn't would you get deducted marks. It only says show the stationary points for 2 marks.

 

[crazy_paki123]

for 14.a.ii. the graph sketching question. Did you have to write the x intercepts when drawing the graph? if you didn't would you get deducted marks. It only says show the stationary points for 2 marks.

no you dont have to show the intercepts only the stationary points