Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
martian86
Group Title
I was trying to find the roots of the polynomial f(x) = 4x^3+3x^26x1.
By Rolle's theorem, If there exists 2 different values of x, say x(1) and x(2) that returns a same value of f(x),
then at some point between x(1) and x(2), d/dx[f(x)] = 0. > (1)
Hence, I differentiate f(x) to get, d/dx[f(x)] = 12x^2+6x6
By (1) 12x^2+6x6 = 0
Solving for x, i get 2 roots, x = 1 and 1/2.
Now solve for f(x) = 4x^3+3x^26x1 with x = 1 (by changing the sign) and surprisingly, f(1) = 0 !!,
seems to be one of the root of f(x).
Can someone help me to understand this ?
 one year ago
 one year ago
martian86 Group Title
I was trying to find the roots of the polynomial f(x) = 4x^3+3x^26x1. By Rolle's theorem, If there exists 2 different values of x, say x(1) and x(2) that returns a same value of f(x), then at some point between x(1) and x(2), d/dx[f(x)] = 0. > (1) Hence, I differentiate f(x) to get, d/dx[f(x)] = 12x^2+6x6 By (1) 12x^2+6x6 = 0 Solving for x, i get 2 roots, x = 1 and 1/2. Now solve for f(x) = 4x^3+3x^26x1 with x = 1 (by changing the sign) and surprisingly, f(1) = 0 !!, seems to be one of the root of f(x). Can someone help me to understand this ?
 one year ago
 one year ago

This Question is Open

vladislav1 Group TitleBest ResponseYou've already chosen the best response.0
I will explain you how to find roots of the f(x). First root we will guess. It will be 1 or 1. It is deviders of the last member of the equation. f(1)=4*(1)^3+3*(1)^26*(1)1=4+3+61=4 f(1)=4*(1)^3+3*(1)^26*(1)1=4+361=0 => x_(1)=1  the root devide column the equation: (4x^3+3x^26x1)/(x1)=4x^2+7x+1 then we find the roots of 4x^2+7x+1, D=7^24*4*1=33 x_(2)=(1+sqrt(33))/8 x_(3)=(1sqrt(33))/8
 one year ago

martian86 Group TitleBest ResponseYou've already chosen the best response.0
Thx for the post, but my question was not about the roots of the equation. If u closely observe this, f'(x) = 12x^2+6x6 has roots one of the roots as 1. and strangely, one of the roots of the equation, f(x) = 4x^3+3x^26x1 is 1 [f(1) = 0] By changing the sign of one of the root of d/dx[f(x)], i actually get one of the real root of f(x). This is cool, but I cant figure out all the cases, where such a relationship can exist !!! Or to put it the other way, why does f(x) has root of 1 and d/dx[f(x)] has root of 1. I some how cant accept that this is just a coincidence..
 one year ago

vladislav1 Group TitleBest ResponseYou've already chosen the best response.0
it is concurrence, look at graphics http://www.yotx.ru/default.aspx?clr0=000000&exp0=4x+%5E+3+%2B3+X+%5E+26x1+&mix=5&max=5&asx=on&u=mm&nx=X&miy=10&may=5&asy=on&ny=Y&iw=600&ih=400&ict=png&aa=on, in x=1 and x=0.5 f(x) have extremums, in x_(1)=1, x_(2)= (1+SQRT(33))/8, x_ (3)=(1SQRT(33))/8 f(x) have roots for example, take another function, g(x)=5x^3+7x^2+2x1, look http://www.yotx.ru/default.aspx?clr0=000000&exp0=5x%5E3%2B7x%5E2%2B2x1&mix=5&max=5&asx=on&u=mm&nx=X&miy=10&may=5&asy=on&ny=Y&iw=600&ih=400&ict=png&aa=on, roots and extremums  the another dotes
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.