Results 1 to 7 of 7

Math Help - Finding the roots of an equation with its derivative?

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    119

    Finding the roots of an equation with its derivative?

    I remember my friend showing me a way to find the max/min points of an equation with its derivative and remember him faintly telling me you can also find the roots with its derivative. I was wondering if this were true and if there are any faster methods than the factor/remainder theorem (guess and check + long division).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    You can by solving \dfrac{dy}{dx} = 0 to find a stationary point. You find if it's a min or max by checking the sign of \dfrac{d^2y}{dx^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    119
    Quote Originally Posted by e^(i*pi) View Post
    You can by solving \dfrac{dy}{dx} = 0 to find a stationary point. You find if it's a min or max by checking the sign of \dfrac{d^2y}{dx^2}
    Thanks, but I was wondering if you could find the roots (zeros) of a function.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    447
    Quote Originally Posted by youngb11 View Post
    Thanks, but I was wondering if you could find the roots (zeros) of a function.
    depends on the function ... could you be more specific?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    119
    Quote Originally Posted by skeeter View Post
    depends on the function ... could you be more specific?
    Say F(x)=x^3-4x^2+x+6

    Derivative would be 3x^2-8x+1. Would I be able to use that to figure out the roots/zeros of a function?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    447
    Quote Originally Posted by youngb11 View Post
    Say F(x)=x^3-4x^2+x+6

    Derivative would be 3x^2-8x+1. Would I be able to use that to figure out the roots/zeros of a function?
    since b^2-4ac = 52 , the quadratic will not factor over the rational numbers ... use the quadratic formula.

    \displaystyle x = \frac{4\pm \sqrt{13}}{3}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    It looks to me like both e^(i pi) and skeeter have misunderstood the question. e^(i pi) talked about finding critical points and skeeter about solving f'(x)= 0. The question, I believe, is about using f'(x) to solve f(x)= 0.

    It sounds to me like you are talking about "Newton's method" for solving equations. The basic idea is that you replace the function with its tangent line approximation and solve the resulting linear equation. Of course, a given graph will have different tangent lines at different points on its graph so you have to select a starting value for x. If we start with x_0 as the starting value, the slope of the tangent line is f '(x_0) and the tangent line approximation to f(x), about the point (x_0, y_0), is y= f '(x_0)(x- x_0)+ f(x_0). Setting that equal to 0, we have f '(x_0)(x- x_0)+ f(x_0)= 0 so f '(x_0)(x- x_0)= -f(x_0), x- x_0= -\frac{f(x_0)}{f '(x_0)}, and, finally, x= x_0- \frac{f(x_0)}{f'(x_0)}. As long as you stay away from values of x_0 that make f '(x_0) close to 0 (so that fraction will not "blow up") that x will be reasonably close to the solution to f(x)= 0. You can make it better by repeating this- choose " x_0" to be this new value of x and repeat- as often as you like.

    Taking your example, let F(x)= x^3- 4x^2+ x+ 6 so that F'(x)= 3x^2- 8x+ 1. Taking x_0= 1, because that is close to the min point of the quadratic so away from F'= 0, we have F(1)= 1- 4+ 1+ 6= 4, F '(1)= 3- 8+ 1= -4 and x= 1- (4/-4)= 2. Now, F(2)= 8- 16+ 2+ 6= 0 and F'(2)= 24- 16+ 1= 9 so that x= 2- (0/9)= 2. Of course, that is because F(2)= 0 so that 2 is, in fact, a solution.

    Suppose the equation had been f(x)= x^2- 2 (so that we are trying to find the square root of 2). Now f'= 2x. Taking x_= 1, we have f(1)= 1- 2= -1 and f'(1)= 2. Then x= 1- (-1/2)= 3/2= 1.5. Taking x_0= 3/2 now, f(3/2)= 9/4- 8/4= 1/4 and f'(3/2)= 3. x= 3/2- ((1/4)/3)= 3/2- 1/12= 18/12- 1/12= 17/12= 1.41666..

    As you can see that is converging pretty rapidly to \sqrt{2}= 1.414.....
    Last edited by HallsofIvy; January 18th 2011 at 07:41 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the roots of this equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: April 11th 2011, 08:10 PM
  2. finding roots of equation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 25th 2010, 01:28 PM
  3. Finding equation from roots
    Posted in the Algebra Forum
    Replies: 6
    Last Post: October 17th 2009, 05:27 AM
  4. Replies: 3
    Last Post: April 29th 2009, 10:09 AM
  5. Finding equation using roots
    Posted in the Algebra Forum
    Replies: 6
    Last Post: July 9th 2007, 12:28 PM

Search Tags


/mathhelpforum @mathhelpforum