Q9c (2 Viewers)

[sinsolja]

I meant your point at 3 , its meant to be a stationary point, so it woud'nt be v-shaped.

 

[mitchh81]

yes this is what i thought ^^

 

[CecilyMare]

someone draw the actual answer pls, i'm not too sure on it but the v shape is ridiculous.

 

[michaeljennings]

Who got this...

View attachment 23396

It's a shitty picture - I know.

At x=3, there should be a minimum turning point, not really sure what you did there.
But you should get 2 marks for f'(x)=0 at x=1, and horizontal asymptote at y=0,

 

[someth1ng]

Is the whole shape more like a parabola? With a stuff right-side?

 

[michaeljennings]

Is the whole shape more like a parabola? With a stuff right-side?

Yes, but after x=3 theres a horizontal asymptote along the x axis.

 

[CecilyMare]

this is what i did
blue line dotted

 

[someth1ng]

I don't understand why there's an inflexion point beetween 1 and 3.

 

[shutupnsmile]

if you cross the x-axis again means you got another stationary point.....now you will never repeat this mistake in your life hahhas

 

[someth1ng]

if you cross the x-axis again means you got another stationary point.....now you will never repeat this mistake in your life hahhas

The one in CecilyMare's graph.

 

[michaeljennings]

I don't understand why there's an inflexion point beetween 1 and 3.

There shouldn't be, it should just go straight down

 

[shutupnsmile]

If the concavity changes for f(x), doesn't that mean that f '(x) will reach positive because the concavity of f(x) has changed????


Do you get what I mean ?? Argh fk it, I probably did it wrong LOL

Yes and No, because they give you the asymtote tells you that the particle is slowing down. i.e. it approaches 0

 

[someth1ng]

Like this?

 

[michaeljennings]

Like this?

Yep, that's it.

 

[shutupnsmile]

Like this?

yup, but im not sure how they mark it, i showed no working only drew it :S

 

[someth1ng]

Well, that explains why mine looked fucked as fuck...LOL

 

[CecilyMare]

don't understand. why is there no inflexion point at 1?

 

[sinsolja]

Because theres a stationary point on F(x) at 1. Stationary point on F(X) at 1 means that F(dash)(X) is 0 at that point.

 

[someth1ng]

I realised why I fucked up - I mixed up that a turning point turned into an inflexion point and 0 - It's obviously just 0.

 

[_pizza]

Is it supposed to be pointed or not?

Not pointed, but I doubt they'd deduct marks for it. I think its just a point of inflexion from memory, so would change concavity without being stationary, and then would never cross the x axis because as f(x) approaches 8, f'(x) approaches 0. If f'(x) crossed the x axis, this would mean the gradient would be positive again, so f(x) would be increasing, and thus would cross the horizontal asymptote of y = 8.

EDIT: I now realise most of this is redundant, as other people have previously given better explanations.