While studying the graph of a cubic function(with real roots) I realized that the line joining the critical points of the function approximately passes through one of the roots. I will demonstrate an example.

Let the cubic function be $f(x) = x^3 - 21x^2 + 138x -280 $. (The roots of this function are $x = 4, x = 10, x = 7$)

On finding the derivative and solving for the critical values, we get the values to be

$a = 5.26$, $ b = 8.73 $

Then the critical points will be $f(5.26)$ and $f(8.73)$

The equation of the line joining these two points will be

$$\frac{y - f(a)}{x - a} = \frac{f(b) - f(a)}{b-a}$$

Putting $y = 0$ and solving for $x$ we get

$x = 6.948 $

which is approximately equal to one of the roots of this fucntion($x = 7$)

I would like to know more about this.

  • 1
    $\begingroup$ You just happened to be lucky. Newton's method is related, try looking that up $\endgroup$ – Simply Beautiful Art Sep 25 '16 at 11:10
  • $\begingroup$ i have found $$x_1=-7-\sqrt{3},x_2=-7+\sqrt{3}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 25 '16 at 11:47
  • $\begingroup$ can you say please what your question exactly is? $\endgroup$ – Dr. Sonnhard Graubner Sep 25 '16 at 11:51
  • $\begingroup$ I suppose that you meant $x^3-21 x^2+138 x-280=0$ $\endgroup$ – Claude Leibovici Sep 25 '16 at 11:52

You have performed one step of the secant method for root-finding. If you perform more steps, you'll get a better approximation of the root.


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