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HellVeN

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Whilst revising, I came across a question where using y'' to find the nature of a stationary pt didn't work. Now I know I could just have found its nature by using by testing y' of LHS and RHS but I was wondering why the method that uses y'' didn't work?
And If it doesn't work iss it safer to always use the testing y' LHS and RHS method?


The question was this:

Find the stationary point on the curve y = (3x^4)+1 and determine what type of point it is.


I did this the y'' way and it doesn't work...
y = (3x^4)+1
y' = 12x^3
y'' = 36x^2
.:. for stationary pts y'=0
12x^3 = 0
x = 0
Now y'' = 36*0^2 = 0
So now I'm thinking this fucker's a possible horizontal pt of inflexion, and i test it and it comes out that LHS is + and RHS is + , therefore concativity doesn't change and I'm fucked.


The answer says it's a minimum and I knew this cause its an x^4 but WHY doesn't the method above work? Shouldn't I have gotten y'' > 0 for a minimum st point?
 

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HellVeN said:
Whilst revising, I came across a question where using y'' to find the nature of a stationary pt didn't work. Now I know I could just have found its nature by using by testing y' of LHS and RHS but I was wondering why the method that uses y'' didn't work?
And If it doesn't work iss it safer to always use the testing y' LHS and RHS method?


The question was this:

Find the stationary point on the curve y = (3x^4)+1 and determine what type of point it is.


I did this the y'' way and it doesn't work...
y = (3x^4)+1
y' = 12x^3
y'' = 36x^2
.:. for stationary pts y'=0
12x^3 = 0
x = 0
Now y'' = 36*0^2 = 0
So now I'm thinking this fucker's a possible horizontal pt of inflexion, and i test it and it comes out that LHS is + and RHS is + , therefore concativity doesn't change and I'm fucked.


The answer says it's a minimum and I knew this cause its an x^4 but WHY doesn't the method above work? Shouldn't I have gotten y'' > 0 for a minimum st point?
I think I know where you went wrong...

You found y by using the 2nd derivative. But to find y you sub x into the original equation, which would mean y=1.

Then again, I said 'I think' :p
 

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yeah, i think uh...the person above this post is right, substitute x into the original equation to find y.
 

wrong_turn

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y' is for the stationary points or possible points of inflexion.
y" is for its nature, that is concave up, down or point of inflexion.
 

acmilan

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HellVeN:

when you go to test it by subbing in a point just before and just after the turning point you will find that the signs do change:

y' = 12x3

the stationary point is at x = 0, so we will test with x = -0.1 and x = +0.1

Clearly, (-0.1)3 is a negative and (0.1)3 is positive, so the LHS of the stationary point is - and the RHS is +, thus giving the turning point, making the graph look similar to an y = x2 + 1 graph would look.
 
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HellVeN

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No no lol, I don't think any of you understand my question :/


I'm aware that there are 2 methods for finding out if a stationary point is a minimum, a maximum and a horizontal point of inflexion.


First we find stationary points by doing y' = 0 (obvious as that means gradient is 0, ie:it's stationary)

The methods to check if the curve is a minimum, a maximum or a horizontal point of inflexion are:

METHOD 1. Check LHS and RHS of f'(x)
METHOD 2. check the nature of f"(stationary_pt)


The thing is that for the equation which I posted earlier, METHOD 2 fails to work which leads me to my original question, WHY?

Acmilan, I know the first method worked for the equation, but I would like to know why the second one doesn't?
 

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Because the concavity must change at the point x=0 on y=x<sup>4</sup> the concavity is neither positive or negative, however it doesn't change hence it isn't a point of inflexion.

Simply getting f''(x)=0 means shit you must also get f''(x-&delta;x)=s and f''(x+&delta;x)=t, where s and t are opposite in sign.

For example you can not state that y=x<sup>3</sup> has a point of inflexion just because f''(0)=0 you must show that f''(-&delta;x)<0, f''(&delta;x)>0.
 

HellVeN

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Hmm...

My book says this:

If f'(x)=0 and f''(x) > 0 , then it's a minimum turning point
If f'(x)=0 and f''(x) < 0 , then it's a maximum turning point
If f'(x)=0 and f''(x) = 0 and concativity changes, then it's a horizontal point of inflexion


So take this simple curve:
y=x^4
y'=4x^3
y''=12x^2
for stationary point, y' = 0
.:. x = 0
stationary point at (0,0)
f''(0) = 0
[ROFL so this tells me that it's not a maximum or a minimum, it tells me it's a fucken possible horizontal point of inflexion, but we obviously know that the stationary point of y=x^4 is a minimum. And this is precisely what I don't understand, according to the rules i gquoted above from my textbook, if it was a minimum it would have been y'' > 0.]

What is it that I'm missing here :(. The way I see it at the moment it seems that using the first derivate only to find the nature of the sp is the only viable option. Is the book wrong or what?
 

HellVeN

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Maths in Focus - 3 Unit mathematics- Book Two - Margaret Grove
 

HellVeN

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Ok according to other topics, it seems that if y'' = 0 then you just have to test it as it could be a minimum / maximum / inflexion.
Stupid textbook.
 

*ashlea*

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yeah, i have that text book 2, u gotta watch it cos there are a few mistakes like that.. it also often uses very different methods to solve problems than what yor class teacher may teach u, especially when u get 2 diff and integration of inverse trig functions.. just ask yor teacher or sum1 with a diff. text if u get confused..
 

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HellVeN said:
Ok according to other topics, it seems that if y'' = 0 then you just have to test it as it could be a minimum / maximum / inflexion.
Stupid textbook.
Alternatively, you could use the third derivative test.
 

HellVeN

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*ashlea* said:
yeah, i have that text book 2, u gotta watch it cos there are a few mistakes like that.. it also often uses very different methods to solve problems than what yor class teacher may teach u, especially when u get 2 diff and integration of inverse trig functions.. just ask yor teacher or sum1 with a diff. text if u get confused..
My teacher is not very good, he makes lots of mistakes, doesn't explain things well and usually doesn't know the answer to questions :/, can't trust em.
 

*ashlea*

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lol.. mine 2! he had a 15min argument with me the other day after i got a different answer 2 him in a general solutions question.. he said i was wrong and he was right.. then when he found the answers and they matched mine, he said theirs were wrong 2! then he went and got the head maths teacher to do it on the board.. he did it the same way as me, got the same answer, and my teacher goes.. but what about this way, isnt this the write answer? and the head goes, "no, you're wrong" and my teacher goes "but, but..." and the head goes "you're wrong ok!?"
lol.. he was so cut.. and embarrassed.. but man it was funny..
 

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