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sghguos

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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

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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... :confused2:

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

RealiseNothing

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

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

iv) k > 32 iirc.
 

Currybear

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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

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

x was increasing for -2<x<0 and >1.
 

sghguos

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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....
 

Timske

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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'
 

Kimyia

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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

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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.
 

megaman64

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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

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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 :)
 

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