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#1 2007-08-14 13:50:48

sezza
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Graphs

The graph y= 3x^4 - 4x^3 - 12x^2 + 36 has two points of inflection.
The x-coordinates of these two points are given by the solution of 3x^2 + bx + c = 0

Find b and c

I know they both equal -2 but how do i get that?

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#2 2007-08-14 19:23:15

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: Graphs

b = -3
c = -6

why it tells you to derive it as 3x^2 + bx + c = 0 when you can reduce it to x^2 + bx + c = 0 is beyond me.

all together there are 3 stationary points: x = 2,x = -1, x = 0

x = 2, x = -1 are local minima, x = 0 is a local maxima.

none of the 3 are points of inflection, so that seems to be another mistake in the question?

Last edited by luca-deltodesco (2007-08-14 19:28:42)


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#3 2007-08-14 21:49:37

mathsyperson
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Registered: 2005-06-22
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Re: Graphs

If I remember, points of inflection are merely points where the second derivative is 0. They don't also have to be stationary.


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#4 2007-08-14 22:28:39

luca-deltodesco
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Re: Graphs

ive always been taught points of inflection had to be stationary points, but incase its not:

dy/dx = 12x^3 - 12x^2 - 24x
d^2y/dx^2 = 36x^2 - 24x - 24 = 0

3x^2 - 2x - 2 = 0

which agrees with with original poster said.

i did a bit of research, and thats right, infleciton doesnt have to be stationary point, but this only happens when a graph is rotated or is implicit, which is why ill have never have came across - ive never done rotation of graphs and nothing on infleciton points on implicit functions.

also, that defintion is a little incomplete (infleciton is where second dervititve is 0) since local maxima and minima can also have a second dervitive of 0.

Last edited by luca-deltodesco (2007-08-14 22:46:21)


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#5 2007-08-14 23:19:50

mathsyperson
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Re: Graphs

No they can't. Maxima have a negative 2nd derivative and minima have a positive one.

If a stationary point has a 2nd derivative of 0 then the graph keeps going in the same direction afterwards, meaning it isn't a local extreme.


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#6 2007-08-14 23:34:46

luca-deltodesco
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Re: Graphs

that isn't correct: an example:

y = x^4
dy/dx = 4x^3
d^2y/dx^2 = 12x^2

at x = 0: dy/dx = 0 and d^2y/dx^2 = 0: but the stationary point x = 0 is a local minimum

this is true for any other graph y = x^(2n) where n is an integer more than or equal to 2

Last edited by luca-deltodesco (2007-08-14 23:39:41)


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#7 2007-08-14 23:40:55

mathsyperson
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Registered: 2005-06-22
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Re: Graphs

Hmm, fair point. I must have been misremembering things.


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#8 2007-08-14 23:44:24

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: Graphs

one thing which does strike me weird is how that can be true?

surely, if the derivitive is 0, and the rate of change of the derivitive is 0, then the graph should just keep going in a straight line along the x axis shouldnt it? although, thats not true is it? because the rate of change of the rate of change of the derivitive isnt 0 tongue lol.


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#9 2007-08-15 01:19:42

Ricky
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Registered: 2005-12-04
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Re: Graphs

You have to remember that derivatives are based upon limits.  So it may certainly be that the rate of change of the derivative is 0 at that point. But as soon as you get one epsilon off, you are back to a non-zero derivative.

Just use x^3 as an example if you need to convince yourself more.


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#10 2007-08-15 01:41:02

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: Graphs

i think my explanation is better.

x^3, the third derivitive is 6, so the second,first derivitives can never stay at 0


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