| Author | Message | ||
| jempie New poster Post Number: 1 |
Someone smart help me please! The biggest problem i have with the coming question is that i don't understand its phrasing: "Consider the fucntion f(x)=2x^3+6x^2-4.5x -13.5. State the roots of this cubic and confim using the remainder theorem. Then, taking the roots two at a time, find the equations of the tangent lines to the average of two of the three roots. Find where the tangent lines at the average of the two roots intersect the curve again. Does this oberservation hold regardless of which two roots you average? State a conjecture concerning the roots of the cubic and the tangent lines at the average value of these roots." Problem is I am dutch and have only studied maths in english since september 2005. Therefore the phrasing of this question is difficult. I read a thread on this https://nrich.maths.org/discus/messages/67613/68454.html?1139934327 , however the person who asked the question here understood what "Then, taking the roots two at a time, find the equations of the tangent lines to the average of two of the three roots. Find where the tangent lines at the average of the two roots intersect the curve again." that part of he question meant and knew how to do I. I don't so can someone please help me with this its urgent!!! Thanks a lot! | ||
| Yatir Halevi Veteran poster Post Number: 475 |
It would be easier if you told us exactly what you didn't understand. I don't want to use in my explanation phrases or words that again you wont understand... Yatir | ||
| jempie New poster Post Number: 2 |
Nice i got someones attantion this fast. I don't get the sentences: Then, taking the roots two at a time, find the equations of the tangent lines to the average of two of the three roots. Find where the tangent lines at the average of the two roots intersect the curve again." I don't understand what they mean with finding the equatiosn of the tangent lines to the average... I know what equations of tangent lines are, but i don't know what "to the average" means. Do you see my problem? Don't hesitate to ask again, ur my only hope! | ||
| Yatir Halevi Veteran poster Post Number: 476 |
An average of two numbers a,b is (a+b)/2 you have to find the tangent line equation on the point (x) that is equal to the average of two of the roots. Yatir - I wont be able to reply in the next few hours so if anybody else is reading this please continue to help jempie if he has any more questions. | ||
| jempie New poster Post Number: 3 |
alright now i get it, thanks yatir!!! you gotta admit they phrased it in a difficult way, nevertheless thanks | ||
| jempie New poster Post Number: 4 |
Ok guys, i've got another question, lets hope i get someone's attention. I need to proof that in cubic functions tangent lines to the averages of two zero values intersect the curve of the cubic function again at the third zero value. To show u how far i have come and to give u an introduction here is a quote from an old thread i found: ..."If we start from the beginning, we have: f(x) = a(x-m)(x-n)(x-p), roots are m, n and p. Without loss of generality, we can take the first two roots, m and n, and average them to give (m+n)/2 which we'll call r. Therefore, at x=r, f(r) = a(r-m)(r-n)(r-p). The gradient of the tangent at r is f'(r), which you did using the product rule, and found that f'(r)=a(r-n)(r-p)+a(r-m)(2r-n-p). Now, if you can show that this tangent hits the x axis at p, then this is a general proof, as we have lost no generality in taking the first two roots as m and n and the third one at p. Note we haven't put any numbers in at all. So we want to find an equation for the y-value of the tangent. The conjecture will be proved true if this y-value is equal to 0 when x=p. Since we know that the y-value of the tangent is f(r) at r, AND we know that the gradient of the tangent is a(r-n)(r-p)+a(r-m)(2r-n-p), then we know that the y-value at any x will be f(r) + (x-r)[a(r-n)(r-p)+a(r-m)(2r-n-p)]. This is because gradient = rise / run; the rise is y - f(r) and the run is x - r, so y - f(r) = gradient * (x-r), so y = f(r) + (x-r)f'(r). Therefore, we have y = a(r-m)(r-n)(r-p) + (x-r)[a(r-n)(r-p)+a(r-m)(2r-n-p)]. "... Therefore we have to substitude p for x and 0 for y. However this equation is not easy to simplify, in fact i don't know how to do this except for that 2r=m+n and therefore: y = a(r-m)(r-n)(r-p) + (x-r)[a(r-n)(r-p)+a(r-m)(m-p)] What i would do normally is expanding brackets, but this in the end did not work (it took me a long time i am telling you). So could someone please help me with this? From here you need to show that y=0 when x=p to complete the proof. | ||
| jempie New poster Post Number: 5 |
anyone? | ||
| Graeme McRae Veteran poster Post Number: 1299 |
jempie, your method is good, so you must have made a mistake expanding the brackets. You should look for that mistake. --Graeme | ||
| jempie Poster Post Number: 6 |
ok, but is this really about expanding the brackets? is there not something else, a trick I don't know or a thing to look out for? you gave me some self confidence though, thanks | ||
| Graeme McRae Veteran poster Post Number: 1300 |
There's an easier way to get there, but it's only slightly easier. The way you've chosen will work. I verified that you haven't made any error so far. You have y = a(r-m)(r-n)(r-p) + (x-r)(a(r-n)(r-p)+a(r-m)(m-p)) Now by substituting p in place of x, and dividing through by a, y/a = (r-m)(r-n)(r-p) + (p-r)((r-n)(r-p)+(r-m)(m-p)) Substituting (m+n)/2 in place of r, you have y/a = ((m+n)/2-m)((m+n)/2-n)((m+n)/2-p) + (p-(m+n)/2)(((m+n)/2-n)((m+n)/2-p)+((m+n)/2-m)(m-p)) which can be simplified a whole lot, and then it becomes fairly obvious that y/a=0. | ||
| jempie Poster Post Number: 7 |
wow, thats looking better, i didn't think of deviding through a, thanks Graeme! I hope i can do it now, otherwise, i'll post again if u don't mind | ||
| jempie Poster Post Number: 8 |
m+n/2-m isn't n/2 right? | ||
| jempie Poster Post Number: 9 |
no of course not, sorry about that last post | ||
| Graeme McRae Veteran poster Post Number: 1301 |
m+n/2-m is n/2, but (m+n)/2-m isn't n/2. The latter can be simplified, however, by noting that (m+n)/2-m is the same as (m+n)/2-(2m)/2, and then combining numerators. | ||
| David Franklin Prolific poster Post Number: 369 |
Remember that if this is IB coursework you'll need to print off a copy of this thread to show your teacher!  | ||
| jempie Poster Post Number: 10 |
i know david, but do you know if i am allowed to search for help in this way? | ||
| Vicky Neale Veteran poster Post Number: 1348 |
It might be best to check with your teacher about that: s/he will know the rules best! (No offence meant to David, of course.) | ||
| AaqibAhmed Veteran poster Post Number: 571 |
Well it seems like you have worked the most of it yourself,and others have confirmed your method and given you small hints along the way,so I think you should be fine! Don't worry! | ||
| AaqibAhmed Veteran poster Post Number: 572 |
Saying that above, go with what Vicky suggested! | ||
| jempie Poster Post Number: 11 |
all right, thanks guys, ima post again if i need help, i asked my teacher | ||
| jempie Poster Post Number: 12 |
ok, i did some simplification but i just can't seem to get this right, this is what i did: y/a = ((m+n)/2-m)((m+n)/2-n)((m+n)/2-p) + (p-(m+n)/2)(((m+n)/2-n)((m+n)/2-p)+((m+n)/2-m)(m-p)) y/a = ((n-m)(m-n)(m+n-2p))/2+((2p-m+n)((m-n)(m+n-2p)+(n-m)(2(m-p)))/2 y/a = ((n-m)(m-n)(m+n-2p)+(2p-m+n)((m+n-2p)+(2(m-p)))/2 y/a = ((n-m)(m-n)(m+n-2p)+(2p-m+n)(3m+n-4p))/2 y/a = ((n-m)(m-n)(m+n-2p)+(2p-m+n)(3m+n-4p))/2 y/a = ((-n^2m-n^3+2pn^2-m^3-nm^2+2pm^2+2nm^2+2mn^2-4pnm)+(2p-m+n)(3m+n-4p))/2 y/a = ((-n^3+2pn^2-m^3+2pm^2+nm^2+mn^2-4pnm)+(2p-m+n)(3m+n-4p))/2 Ok and that last bit clearly isn't equal to zero, so somewhere there is a mistake. However my knowledge, time and self esteem is lacking. Therefore please help me! | ||
| Yatir Halevi Veteran poster Post Number: 477 |
You have some mistakes in your first row. ((m+n)/2-m)((m+n)/2-n)((m+n)/2-p)= -((n-m)(m-n)(m+n-2p))/2 (you forgot a minus sign) and: (p-(m+n)/2)(((m+n)/2-n)((m+n)/2-p)+((m+n)/2-m)(m-p)) isn't equal to what you say it is....check again try doing it slowly and step by step... Yatir | ||
| Graeme McRae Veteran poster Post Number: 1305 |
jempie, (m+n)/2-m is (n-m)/2, etc., so as Yatir pointed out, your first two lines should have been y/a = ((m+n)/2-m)((m+n)/2-n)((m+n)/2-p) + (p-(m+n)/2)(((m+n)/2-n)((m+n)/2-p)+((m+n)/2-m)(m-p)) y/a = ((n-m)/2)((m-n)/2)((m+n-2p)/2) + ((2p-m-n)/2)(((m-n)/2)((m+n-2p)/2)+((n-m)/2)(m-p)) Continuing on, the best way to simplify the RHS is to get rid of those pesky factors of 1/2, so 8y/a = (n-m)(m-n)(m+n-2p) + (2p-m-n)((m-n)(m+n-2p)+2(n-m)(m-p)) From there, the rest should be easy. Tedious, but easy. --Graeme | ||
| jempie Poster Post Number: 13 |
ok, thanks, i'll try again, u guys are heros | ||
| jempie Poster Post Number: 14 |
i did it!!! thanks guys | ||
| Graeme McRae Veteran poster Post Number: 1310 |
Jempie, Now that you've done all that tedious work, you should feel proud that you solved the problem with just minor hints. I should point out, as I suggested in my post number 1300, that there's a slightly easier way to go about this. The difference lies mostly in the order in which things are simplified, so they don't get too far out of hand. Apart from that, the method is the same as that you came up with. --Graeme | ||
| jempie Poster Post Number: 15 |
I must confess i don't really feel proud, because 1 i asked lots of stupid questions and two i completely read of the part where you said that there was a slightly easier way to do it. However i am happy i solved it and also that i found a site with a lot of great and helpful people. | ||
| Yatir Halevi Veteran poster Post Number: 482 |
Jempie - Don't worry intuition especially in factorization comes with exprience. You learn to spot patterns... Yatir |