Help: Maths: Applications of Derivatives.. (1 Viewer)

[imoO]

I am absolutely stumped on what to do....

Like I've got a stragtforward question and know what to do, but to work backwards, I've no idea:

This is the question:

Show that y = x^3 - 3x^2 - 6x - 2 has an inflexion point at A (1, -10). A line through A cuts the curve again at B and C. Show that AB = AC

thanks for any suggestions / help in advance,

imoO

 

[foram]

y = x^3 - 3x^2 - 6x - 2
y' = 3x^2 - 6x - 6
y'' = 6x - 6

at x=1 , y''= 0
at x=1+, y''>0
at x=1-, y''<0
Therefore there is a POI at x=1

for
y+10 = (x-1)^3 - 3(x-1)^2 - 6(x-1) -2
y= (x-1)^3 - 3(x-1)^2 - 6(x-1) -12
= x^3 - 3x^2 + 3x - 1 - 3x^2 +6x -3 - 6x + 6 - 12
= x^3 - 6x^2 + 3x - 16
= f(x)

f(-x) = (-x)^3 - 3x^2 + 9(-x) - 16
-f(-x) = (x)^3 +3x^2 + 9(x) +16

f(x) does not equal -f(-x)?

I can't do this either...

I was thinking that if the curve was odd around the point of inflection, then AB is a rotation of AC... but it doesn't look like it's works out that way.

 

[kaz1]

you started calculus already were way behind then

 

[Continuum]

Nah, some schools do that first.

 

[jellybelly59]

lol i got the inflexion bit... but den just got lost after that hahaha

 

[Devouree]

good work foram! I was going to do something like that, but what if you use the third derivative?

if f''(a)=0, f'''(a) *doesn't equal* 0, then it's an inflexion point.

If your just going on about the second derivative, then it MIGHT be a POI, but you can't be certain. And then to find out where the point is you can just sub it in.

LHS=y, RHS= *sub x into here...*

(can't actually be stuffed to do the question, but hope that helps.

 

[lyounamu]

I am absolutely stumped on what to do....

Like I've got a stragtforward question and know what to do, but to work backwards, I've no idea:

This is the question:

Show that y = x^3 - 3x^2 - 6x - 2 has an inflexion point at A (1, -10). A line through A cuts the curve again at B and C. Show that AB = AC

thanks for any suggestions / help in advance,

imoO

Ok, the first one is already done by Foram so I won't answer that. For the second one, you need to find the equation of the line that cuts A (by using y-y1 = m(x-x1)). Second one is quite easy too.

You can find the gradient by finding first derivative and substitute the 1 into the x-value.

After you find the equation, you use the simultaneous equation (of the line that cuts A and y= x^3 - 3x^2 - 6x -2) to find the coordinates of B & C because B and C are coordinates where these two equations meet.

Then use the distance formula and prove that AB = AC.

I would like to show you all the working out but I am running out of time as I need to go to school. I will get you the full working as soon as I get back. Sorry!

EDIT: I found that the equation does not cross the equation again. Therefore, B & C do not exist...

EDIT 2: My solution only works when the equation is the tangent to the point A. I have got limited info on the equation of A. It is impossible to find the points B & C unless you know what the equation of A is.

FURTHER EDITION: ONLY WAY TO FIND THE ANSWER is to substitute any gradient (e.g.-3) to find the equation of A and then use the simultaneous equation to find the coordinates of B & C. Then use the distance formula to prove that AB = AC. The question is extremely ambiguous.

 

[foram]

good work foram! I was going to do something like that, but what if you use the third derivative?

if f''(a)=0, f'''(a) *doesn't equal* 0, then it's an inflexion point.

If your just going on about the second derivative, then it MIGHT be a POI, but you can't be certain. And then to find out where the point is you can just sub it in.

LHS=y, RHS= *sub x into here...*

(can't actually be stuffed to do the question, but hope that helps.

I used the 2nd derivative test for the Point of inflection. I don't need to find the 3rd derivative, it's just more work.

 

[lyounamu]

good work foram! I was going to do something like that, but what if you use the third derivative?

if f''(a)=0, f'''(a) *doesn't equal* 0, then it's an inflexion point.

If your just going on about the second derivative, then it MIGHT be a POI, but you can't be certain. And then to find out where the point is you can just sub it in.

LHS=y, RHS= *sub x into here...*

(can't actually be stuffed to do the question, but hope that helps.

You don't go as far as trying to find the third derivative. There is no need to and that is not even required. To test the point of inflexion, only thing that is required is 2nd derivative.

 

[Devouree]

Oh sorry, I was taught that with the second derivative you can't conclusively prove that its a POI, only if you find the third derivative and it isn't 0 can you say that its a POI. (mebbe wrong...)

 

[foram]

Oh sorry, I was taught that with the second derivative you can't conclusively prove that its a POI, only if you find the third derivative and it isn't 0 can you say that its a POI. (mebbe wrong...)

You only need to show that it changes concavity on either side of where the second derivative is equal to 0.

It is only inconclusive if you show that the second derivative is equal to 0 without showing the change in concavity.

If you use the 3rd derivative, you may need to differentiate something horrible, and that could cost a lot of time. Thats why most people just test using the 2nd derivative.

 

[tommykins]

Second derivative is all that's required.

 

[imoO]

ROFL...we didn't have to do that question...ahaha!!...lol...I just found out today....anyway...I have another one I'm not too sure of.

The curve y=ax^2+bx+c has a maximum point at (2,18) and it passes through the point (0,10). Find the values of a, b and c.

For whoever said they were behind...no not really...I'm self taught because my school is shit, nobody has heard of it, and we use Maths-In-Focus which is typically a shit book for 3-u as everybody says. Getting my hands on a Cambridge is hard, I've been looking and haven't had any luck so far. I live in Sydney, if anybody has ideas, please tell me.

 

[SoulSearcher]

ROFL...we didn't have to do that question...ahaha!!...lol...I just found out today....anyway...I have another one I'm not too sure of.

The curve y=ax^2+bx+c has a maximum point at (2,18) and it passes through the point (0,10). Find the values of a, b and c.

For whoever said they were behind...no not really...I'm self taught because my school is shit, nobody has heard of it, and we use Maths-In-Focus which is typically a shit book for 3-u as everybody says. Getting my hands on a Cambridge is hard, I've been looking and haven't had any luck so far. I live in Sydney, if anybody has ideas, please tell me.

Find the first derivative, then sub the points into the equation of the parabola and simultaneously solve. Remember that at a turning point, the first derivative is equivalent to 0. You should get a = -2, b = 8 and c = 10, if I thought it through right.

 

[lyounamu]

If the curve passes through (0,10),
10 = a . 0^2 + b . 0 + c
Therefore, c= 10

Now, the curve has the maximum point at (2,18),
Therefore, 18 = a . 2^2 + b .2 + 10
18 = 4a +2b +10
8 = 4a + 2b ...(1)
y=ax^2 + bx + c
y'=2ax + b

The curve has the maximum point at (2,18)
Therefore, 0 = 2 x 2 x a + b
0 = 4a +b ...(2)

(1) - (2):
8 = b

Substitue 8 as the value of b into the either equation (1) or (2).
8 = 4a + 16
4a = -8
a = -2

I suggest you use Cambridge. Another great book to recommend is Fitzpatrick. It is my primary textbook and I think that book is one of the best you can use.

 

[imoO]

thanks