Question 5.... (1 Viewer)

[Beaky]

Im just worried about this one...

FOr the two stationary points did people get a

Horizontal Inflection at (0,0) and Mininum at (3, -27)

I choked on this question- i nearly cried coz i can do anything like this...

 

[Belle]

Yep, that's what I got.

Fingers crossed that we're right!

 

[swangal]

yeh thats wat i got-exactly the same answer- but i dunno if u can trust my dodgy answers!

 

[shazabdazla]

yep i got the same...i think were al right...

 

[SmokedSalmon]

yep same here

 

[iambored]

Originally posted by Beaky
Im just worried about this one...

FOr the two stationary points did people get a

Horizontal Inflection at (0,0) and Mininum at (3, -27)

I choked on this question- i nearly cried coz i can do anything like this...

me too, and i also got another point of inflexion around (1,3) (you have to text y''=0)
it was question 5 and i seriously think that they picked the hardest they could from each topic. like, they couldnt have a normal max and min could they, no it was 2 points of inflexion and a min!! argh!!!

 

[sneaky pete]

There was also a point of inflexion when x = 2 right?
I included that but since the question only asked to sketch the curve showing stationary points, for 1 mark, I dont think it matters?

I got the same answers as above, they should be ok.

 

[ace]

what does it mean in q5(iv.) for
find the values of x for which the graph of y=f(x) is concave down. What does that mean, also what is the answer for that.

 

[CHUDYMASTER]

That inflection point is needed for part (iii).

You needed to show where it was concave down and it's:
0< x < 2 since it starts to concave up past that inflection point.

 

[iambored]

Originally posted by iambored
me too, and i also got another point of inflexion around (1,3)

alright, around (1,2) then, can't remember

you can do y''<0, because that is when it it concave down, that's when it's a max and hence is facing downwards

then you solve that out, and ger that chudymaster got

 

[Beaky]

It should be 0 < x < 1

Cause point of inflexion is when f''(x) = 0

So 12x^2-12x = 0
x= 1

Am I correct???

 

[Adam]

y'' doesn't equal 12x^2-12x does it??

But yeah, 3, -27 for the max.

 

[Kn1ght_M4r3]

y' = 4x^3 - 12x^2
y" = 12x^2 - 24x

 

[Kn1ght_M4r3]

y' = 4x^3 - 12x^2
y" = 12x^2 - 24x

 

[cakes]

yay i got something like that!! (which probably means you guys are wrong ) lol

 

[Winston]

for iv for the x values that make it concave down i got this


-2<x<2


is that right?

 

[braindrainedAsh]

Originally posted by Beaky
Im just worried about this one...

FOr the two stationary points did people get a

Horizontal Inflection at (0,0) and Mininum at (3, -27)

Yep I got that too.

 

[Bandit]

Yeah i got that... what a relief

 

[dnt stop runnin]

the curve is concave down between 0<=x<=2.....bcause of the y''.......thats wat i rekon??.....wat u all think???

 

[tempco]

it stops being concave down at the other point of inflexion..can't remember what it was... the curve had 2 points of inflexion.