Q9c (2 Viewers)

sinsolja

Member
Joined
May 16, 2009
Messages
86
Gender
Undisclosed
HSC
2012
I meant your point at 3 , its meant to be a stationary point, so it woud'nt be v-shaped.
 

shutupnsmile

Premium Member
Joined
Sep 9, 2008
Messages
49
Gender
Undisclosed
HSC
2009
if you cross the x-axis again means you got another stationary point.....now you will never repeat this mistake in your life hahhas
 

shutupnsmile

Premium Member
Joined
Sep 9, 2008
Messages
49
Gender
Undisclosed
HSC
2009
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
 

sinsolja

Member
Joined
May 16, 2009
Messages
86
Gender
Undisclosed
HSC
2012
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

Retired Nov '14
Joined
Sep 18, 2010
Messages
5,558
Location
Adelaide, Australia
Gender
Male
HSC
2012
Uni Grad
2021
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

Member
Joined
Mar 31, 2011
Messages
37
Location
Space
Gender
Male
HSC
N/A
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.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top