Stationary Points, Asymptotes and Points of Inflexion. (2 Viewers)

Michaelmoo · Feb 9, 2009

Ok Basically the question says

Find any Stationary points, Points of Inflexion or asymptotes for:

y = 2x + 3
. . .(x^2 - 4)

Ok so I find the asymptotes, X = plus or minus 2 and y = 0

When I try to find any stationary points or Points of inflexion, I get no solution. Although Clearly, there should be a horizontal point of inflexion at (0, -3/4), but I can't get this.

This means that the first and second derivative must equal zero at this point.

Could someone please run me through this question??

Thanks in advance.

skyline · Feb 9, 2009

Michaelmoo said:
Ok Basically the question says

Find any Stationary points, Points of Inflexion or asymptotes for:

y = 2x + 3
. . .(x^2 - 4)

Ok so I find the asymptotes, X = plus or minus 2 and y = 0

When I try to find any stationary points or Points of inflexion, I get no solution. Although Clearly, there should be a horizontal point of inflexion at (0, -3/4), but I can't get this.

This means that the first and second derivative must equal zero at this point.

Could someone please run me through this question??

Thanks in advance.

its easy mate

for stat point differentiate to first derivative, you need to use your quotient rule and equate it to zero then what ever you get subs in the original equation and find the y solution

point of inflexion is easier, you just do second derivative of the function and equate to zero then subs your value into the first derivative equation, i think, either that or the original i cant remember for sure, im saying this on the top of my head

but your asymptotes appear right its basically domain and range values like you said

try that and see if you can get a solution, hope my advice helps

cheers.

Michaelmoo · Feb 9, 2009

skyline said:
its easy mate

for stat point differentiate to first derivative, you need to use your quotient rule and equate it to zero then what ever you get subs in the original equation and find the y solution

point of inflexion is easier, you just do second derivative of the function and equate to zero then subs your value into the first derivative equation, i think, either that or the original i cant remember for sure, im saying this on the top of my head

but your asymptotes appear right its basically domain and range values like you said

try that and see if you can get a solution, hope my advice helps

cheers.

Thanks, but that doesn't solve te problem. See I know how to do it but I'm not getting the answer that I should. I;ve reviewed my question and no mathematical error either.

Try it yourself and see what you can get out of it.. I'm gettign no stats or points of inflexion.

O and BTW, it's the original equation, not the first derivative. Although initially you have to check for a change in concavity by testing either side of the possible POI, by subbing it into the 2nd derivative.

addikaye03 · Feb 9, 2009

there is no stat points, but inflection at x= -1.5.. i didnt attempt the Q just typed it into graphmatica program. If no has done it, later tonight i will do it for you

EDIT: ok, ill do it now lol give me a sec

skyline · Feb 9, 2009

well there you go, easier said then done, appears right to me

i wasn't solving the problem for you mate, i was just guiding you roughly through it, i could solve it, but i was too lazy, and i was about to do it for you, but that guy above me beat me to it.

and yeah i keep forgetting for inflexion once u found second derivative you sub in the original, thanks for reminding me that.

addikaye03 · Feb 9, 2009

skyline said:
well there you go, easier said then done, appears right to me

i wasn't solving the problem for you mate, i was just guiding you roughly through it, i could solve it, but i was too lazy, and i was about to do it for you, but that guy above me beat me to it.

and yeah i keep forgetting for inflexion once u found second derivative you sub in the original, thanks for reminding me that.

Yea, i sometimes think bout giving guides, but if my solutions come with explanation then i think thats ok. And yea, basically ur just find the y-coordinate to the x value so sub into f(x). And yea its write, i typed the equation into graphmatica to check.

Michaelmoo · Feb 9, 2009

FOr the point of inflexion part:

You've considered the first derivative u'v - vu' but you forgot the v^2 on the denominator?

Also, howcome when you went to find the seond derivative, you didn't consider the denominator of the first derivative (v^2). Also, you didn't use the quotient rule?

Trebla · Feb 9, 2009

$y=\dfrac{2x+3}{x^2-4}=\dfrac{2(x+2)-1}{(x+2)(x-2)}=\dfrac{2}{(x-2)}-\dfrac{1}{x^2-4}\\\dfrac{dy}{dx} = -\dfrac{2}{(x-2)^2}+\dfrac{2x}{(x^2-4)^2}\\$Solve $\dfrac{dy}{dx}=0\\\dfrac{2x}{(x^2-4)^2}=\dfrac{2}{(x-2)^2}\Rightarrow\dfrac{x}{(x+2)^2}=1\\x^2+3x+4=0\,\,\,\,\Delta = -7 < 0 \Rightarrow $no real solution hence no stationary points$\\\dfrac{d^2y}{dx^2}=\dfrac{4}{(x-2)^3}+\dfrac{2(x^2-4)^2-8x^2(x^2-4)}{(x^2-4)^4}=\dfrac{4}{(x-2)^3}+\dfrac{2(x^2-4)-8x^2}{(x^2-4)^3}\\$Solve $\dfrac{d^2y}{dx^2}=0\\\dfrac{4}{(x-2)^3}=\dfrac{8x^2-2(x^2-4)}{(x^2-4)^3}\Rightarrow \dfrac{8x^2-2(x^2-4)}{(x+2)^3}=4\\6x^2+8=4(x^3+3(2)x^2+3(2^2)x+8)\Rightarrow 4x^3+24x^2+48x+32=0$

*sigh* that was messy, might've made a mistake somewhere so check it for me....

Michaelmoo · Feb 10, 2009

Trebla said:
$y=\dfrac{2x+3}{x^2-4}=\dfrac{2(x+2)-1}{(x+2)(x-2)}=\dfrac{2}{(x-2)}-\dfrac{1}{x^2-4}\\\dfrac{dy}{dx} = -\dfrac{2}{(x-2)^2}+\dfrac{2x}{(x^2-4)^2}\\$Solve $\dfrac{dy}{dx}=0\\\dfrac{2x}{(x^2-4)^2}=\dfrac{2}{(x-2)^2}\Rightarrow\dfrac{x}{(x+2)^2}=1\\x^2+3x+4=0\,\,\,\,\Delta = -7 < 0 \Rightarrow $no real solution hence no stationary points$\\\dfrac{d^2y}{dx^2}=\dfrac{4}{(x-2)^3}+\dfrac{2(x^2-4)^2-8x^2(x^2-4)}{(x^2-4)^4}=\dfrac{4}{(x-2)^3}+\dfrac{2(x^2-4)-8x^2}{(x^2-4)^3}\\$Solve $\dfrac{d^2y}{dx^2}=0\\\dfrac{4}{(x-2)^3}=\dfrac{8x^2-2(x^2-4)}{(x^2-4)^3}\Rightarrow \dfrac{8x^2-2(x^2-4)}{(x+2)^3}=4\\6x^2+8=x^3+3(2)x^2+3(2^2)x+8\Rightarrow x^3+12x=0\\x(x^2+12)=0\Rightarrow x=0\\$Test for change in concavity and you have point of inflexion at $x$ = 0$$

*sigh* that was messy, might've made a mistake somewhere so check it for me....

Wow thanks. Erm the second last line line. I think you forgot to multiply through by four on the RHS, in that case there'd be no solution for the second derivative.

I don't know why, but there must be something wrong with this question. Although when you plot it, there should be a horizontal POI at (0,-0.75).

O and btw, howcome you took a split fraction approach, wouldn't it make thinks for complicated?

Trebla · Feb 10, 2009

First of all there is no "horizontal" point of inflexion. A horizontal point of inflexion must satisfy the first and second derivative at zero with change in concavity. A general point of inflexion only needs to satisfy the second derivative and demonstrate a change in concavity on either side.

Secondly, if you look at the graph closely, the inflexion point does NOT happen at x = 0. It happens somewhere between 0 and -1. This is a solution of $4x^3+18x^2+48x+24=0$ which is not a very nice number by the looks of it. I'm pretty sure my second derivative is correct now after the correction, it's just a matter of solving $4x^3+18x^2+48x+24=0$ .

Finally, I took a partial fraction approach to simplify differentiating the first derivative to get the second derivative. If you took the first derivative by the quotient rule without manipulating the fraction you get:
$\dfrac{dy}{dx}=\dfrac{-2x^2-6x-8}{(x^2-4)^2}$
which is slightly more nasty to differentiate and simplify by the quotient rule as opposed to
$\dfrac{dy}{dx} = -\dfrac{2}{(x-2)^2}+\dfrac{2x}{(x^2-4)^2}$
where the second term is much nicer to apply the quotient rule and expand.

addikaye03 · Feb 10, 2009

Michaelmoo said:
FOr the point of inflexion part:

You've considered the first derivative u'v - vu' but you forgot the v^2 on the denominator?

Also, howcome when you went to find the seond derivative, you didn't consider the denominator of the first derivative (v^2). Also, you didn't use the quotient rule?

well i didnt need to consider the denominator cos the numerator=0, 0 divided by anything=0. And yer i see where i fucked up, consequently not using the quotient for the 2nd derivative.

Michaelmoo · Feb 10, 2009

Trebla said:
First of all there is no "horizontal" point of inflexion. A horizontal point of inflexion must satisfy the first and second derivative at zero with change in concavity. A general point of inflexion only needs to satisfy the second derivative and demonstrate a change in concavity on either side.

Secondly, if you look at the graph closely, the inflexion point does NOT happen at x = 0. It happens somewhere between 0 and -1. This is a solution of $4x^3+18x^2+48x+24=0$ which is not a very nice number by the looks of it. I'm pretty sure my second derivative is correct now after the correction, it's just a matter of solving $4x^3+18x^2+48x+24=0$ .

Finally, I took a partial fraction approach to simplify differentiating the first derivative to get the second derivative. If you took the first derivative by the quotient rule without manipulating the fraction you get:
$\dfrac{dy}{dx}=\dfrac{-2x^2-6x-8}{(x^2-4)^2}$
which is slightly more nasty to differentiate and simplify by the quotient rule as opposed to
$\dfrac{dy}{dx} = -\dfrac{2}{(x-2)^2}+\dfrac{2x}{(x^2-4)^2}$
where the second term is much nicer to apply the quotient rule and expand.

Ok. But with this: $4x^3+18x^2+48x+24=0$ , how would you go about solving that? I can't seem to find an integral solution... I don't think there are any solutions....

Also, when I try graphing the original question by testign random points, I see that there should clearly be a horizontal POI at x =0, although It doesn't seem to be working.

O and btw, even the answers say that theres a POI there, although it's out of Maths in focus, it's most likely right.

Trebla · Feb 10, 2009

Michaelmoo said:
Ok. But with this: $4x^3+18x^2+48x+24=0$ , how would you go about solving that? I can't seem to find an integral solution... I don't think there are any solutions....

Also, when I try graphing the original question by testign random points, I see that there should clearly be a horizontal POI at x =0, although It doesn't seem to be working.

O and btw, even the answers say that theres a POI there, although it's out of Maths in focus, it's most likely right.

There is ONE real solution to $4x^3+18x^2+48x+24=0$ , but it's not an integer solution. Just because there is no integer solution doesn't mean there is no solution. It lies between 0 and -1. A graph of $y=4x^3+18x^2+48x+24=0$ should indicate that.

Do you have graphmatica? Plot $y=\dfrac{2x+3}{x^2-4}$ on graphmatica and then plot its derivative dy/dx (use d/dx button near the top right), you will find that the turning point of its derivative (where the inflexion point should occur) is NOT at x = 0. Now plot $y=4x^3+18x^2+48x+24=0$ and you will find the root of the cubic polynomial corresponds to the x-value of the turning point of the dy/dx graph. I am very very sure the inflexion point does NOT happen at x = 0. It happens close to 0 but it's in fact between 0 and -1.

Trebla · Feb 10, 2009

Also, I suggest you try this online differentiator. Enter the original function and then enter its derivative to get the second derivative. It should be clear that x = 0 is NOT a solution to the second derivative.
QuickMath Automatic Math Solutions

Michaelmoo · Feb 10, 2009

Trebla said:
There is ONE real solution to $4x^3+18x^2+48x+24=0$ , but it's not an integer solution. Just because there is no integer solution doesn't mean there is no solution. It lies between 0 and -1. A graph of $y=4x^3+18x^2+48x+24=0$ should indicate that.

Do you have graphmatica? Plot $y=\dfrac{2x+3}{x^2-4}$ on graphmatica and then plot its derivative dy/dx (use d/dx button near the top right), you will find that the turning point of its derivative (where the inflexion point should occur) is NOT at x = 0. Now plot $y=4x^3+18x^2+48x+24=0$ and you will find the root of the cubic polynomial corresponds to the x-value of the turning point of the dy/dx graph. I am very very sure the inflexion point does NOT happen at x = 0. It happens close to 0 but it's in fact between 0 and -1.

Ah ok Thanks. But is there ay particular way to solve that second derivative?

Also, yeah I've graphed it on graphamatica, it's not on x=0 but still looks like a horizontal POI, howcome the first derivative didn't also equal 0 then?

Trebla · Feb 10, 2009

Like I said in the earlier post, it is NOT a horizontal point of inflexion. It is just a point of inflexion. Do you understand the difference?

A point of inflexion satisfies:
- d²y/dx² = 0
- change in concavity on either side of the solution

A HORIZONTAL point of inflexion satisfies:
- dy/dx = 0
- d²y/dx² = 0
- change in concavity on either side of the solution

A horizontal point of inflexion is a special point of inflexion where the curve would flatten out at that point.

In this case, we have just a normal point of inflexion, not a horizontal one.

Michaelmoo · Feb 11, 2009

Trebla said:
Like I said in the earlier post, it is NOT a horizontal point of inflexion. It is just a point of inflexion. Do you understand the difference?

A point of inflexion satisfies:
- d²y/dx² = 0
- change in concavity on either side of the solution

A HORIZONTAL point of inflexion satisfies:
- dy/dx = 0
- d²y/dx² = 0
- change in concavity on either side of the solution

A horizontal point of inflexion is a special point of inflexion where the curve would flatten out at that point.

In this case, we have just a normal point of inflexion, not a horizontal one.

Ah ok. But how would you solve that equation. I mean it does say show any points of inflexions in the original equation? So how would I find the point? Obviously the solution is not integral, but how owuld you find the exact value (or close to it)?

addikaye03 · Feb 11, 2009

Michaelmoo said:
Ah ok. But how would you solve that equation. I mean it does say show any points of inflexions in the original equation? So how would I find the point? Obviously the solution is not integral, but how owuld you find the exact value (or close to it)?

U find the point of inflection by setting the second derivative equal to 0, simple as that! Then ofcourse u go on and check concavity to make sure it is inflection.

Michaelmoo · Feb 11, 2009

addikaye03 said:
U find the point of inflection by setting the second derivative equal to 0, simple as that! Then ofcourse u go on and check concavity to make sure it is inflection.

I understand that. Although, you try setting $y''=4x^3+18x^2+48x+24$ to 0 and solve for x... Is there any possible way?

addikaye03 · Feb 11, 2009

Michaelmoo said:
I understand that. Although, you try setting $y''=4x^3+18x^2+48x+24$ to 0 and solve for x... Is there any possible way?

Graphmatica believes one of the solutions is x=~-0.63, im unsure how do find this value though..put it this way u wouldnt get a Q like this in the HSC if ur required to factorise such a shit expression.

Stationary Points, Asymptotes and Points of Inflexion. (2 Viewers)

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