MedVision ad

geo calculus question (1 Viewer)

darshil

Replicant
Joined
Mar 20, 2008
Messages
234
Location
my mind
Gender
Undisclosed
HSC
2009
Find the stationary points of each of the following curves and use the second derivative to determine their nature
(b) y=1-x^3

I diffrentiated it and made it equal to zero for the stationary points but im stuck at it
i get all the other types, its just this one

i bet its way simple and i cant click on it

thanks heaps !
 

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,250
Gender
Undisclosed
HSC
N/A
Well firstly, look at the equation it's a cubic with it's inflexion (or "vertex" if you like) at (0,1) and it's concavity is opposite to the basic cubic equation (ie. y=x3), So that should give you a hint of what to expect, etc.

y=1-x3
y'=-3x2
y''=-6x

For a stat point y'=0
-3x2=0
x=0

When x=0 y''=0 => No new information. Use the 1st deriv sign test.

When x=0- y'<0
When x=0+ y'<0 [Would ideally use a table for this]

Code:
\__
   \
Therefore horizontal point of inflexion at (0,1)
 

kurt.physics

Member
Joined
Jun 16, 2007
Messages
840
Gender
Undisclosed
HSC
N/A
Find the stationary points of each of the following curves and use the second derivative to determine their nature
(b) y=1-x^3

I diffrentiated it and made it equal to zero for the stationary points but im stuck at it
i get all the other types, its just this one

i bet its way simple and i cant click on it

thanks heaps !
You just need to remember the types of stationary (turning) points







 

sinophile

Well-Known Member
Joined
Oct 25, 2008
Messages
1,339
Gender
Undisclosed
HSC
N/A
Find the stationary points of each of the following curves and use the second derivative to determine their nature
(b) y=1-x^3

I diffrentiated it and made it equal to zero for the stationary points but im stuck at it
i get all the other types, its just this one

i bet its way simple and i cant click on it

thanks heaps !
For stat. pts, let first deriative equal to zero and solve for x. nature is determined by second derivative: sign determines concavity. if it tests as zeo in the 2nd derivative, tetsing of immediate region needs to happen because you canot be sure of its nature atm.

y=1-x^3

y'=-3x^2

y''=-6x

if y'=0

-3x^2=0
x=0

at x=0
f(x)=1
f''(x)=0

test region aroound x=0:
x -1 0 1
y'' + 0 -

therefore 0,0 is a pt of inflexion


edit: i see i did a silly mistake. fuck
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top