A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 3 years ago

I was trying to find the roots of the polynomial f(x) = 4x^3+3x^2-6x-1. 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+6x-6 By (1) 12x^2+6x-6 = 0 Solving for x, i get 2 roots, x = -1 and 1/2. Now solve for f(x) = 4x^3+3x^2-6x-1 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 ?

  • This Question is Open
  1. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)^2-6*(-1)-1=-4+3+6-1=4 f(1)=4*(1)^3+3*(1)^2-6*(1)-1=4+3-6-1=0 => x_(1)=1 - the root devide column the equation: (4x^3+3x^2-6x-1)/(x-1)=4x^2+7x+1 then we find the roots of 4x^2+7x+1, D=7^2-4*4*1=33 x_(2)=(-1+sqrt(33))/8 x_(3)=(-1-sqrt(33))/8

  2. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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+6x-6 has roots one of the roots as -1. and strangely, one of the roots of the equation, f(x) = 4x^3+3x^2-6x-1 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..

  3. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is concurrence, look at graphics http://www.yotx.ru/default.aspx?clr0=000000&exp0=4x+%5E+3+%2B3+X+%5E+2-6x-1+&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)=(1-SQRT(33))/8 f(x) have roots for example, take another function, g(x)=5x^3+7x^2+2x-1, look http://www.yotx.ru/default.aspx?clr0=000000&exp0=5x%5E3%2B7x%5E2%2B2x-1&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

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.