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2008 Mathematics Paper Questions (1 Viewer)

imoO

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2009
Hey guys, just a few questions I would like to request solutions for, because I've been unable to to do them Here is the paper - Download 2008HSC-mathematics.pdf for free on uploading.com Questions- 5 a) 8 a) iii)how do prove it's local maximum and not global? 9 c) ii) 10 a)no clue on how to do the whole question...i think im meant to differentiate in part ii), but that turned out weird. cheers
 
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tonyharrison

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2009
Well for (8)(a)(iii):
f'(x) = 4x^3 - 16x
Stationary points at f'(x) = 0
0 = 4x^3 - 16x
0 = x (4x^2 - 16)
Therefore:
x = 0
OR
4x^2 = 16
x = 2

At (x = 0):
f'(x) = 0 (possible point of inflexion)
Do concavity check, i couldn't be fucked but you can probs assume it is one
At (x = 2)
f'(x) = -16
f'(x) < 0 (maximum turning point)

The endddd!!!
 

daniieee

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2009
Well for (8)(a)(iii):
f'(x) = 4x^3 - 16x
Stationary points at f'(x) = 0
0 = 4x^3 - 16x
0 = x (4x^2 - 16)
Therefore:
x = 0
OR
4x^2 = 16
x = 2

At (x = 0):
f'(x) = 0 (possible point of inflexion)
Do concavity check, i couldn't be fucked but you can probs assume it is one
At (x = 2)
f'(x) = -16
f'(x) < 0 (maximum turning point)

The endddd!!!
You should have gotten 3 stationary points, with one maximum at (0,0) and two minimum turning points at (2,-16) and (-2,-16). You made an error shortly after f'(x)=0 as 4x^3-16x would have given you three x values (being a cubic). From your working you took out the x but neglected to take out the 4 as well which would've made it clearer with 4x(x^2-4) as x^2-4 gives you plus & minus 2.

The link that x_symphonic provided should have the full working.
 

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