we know, real roots are governed by the intersection of any real function with the \(x\)-axis. But, how do we know weather this intersected function with the x-axis at point \(x_{0}\) has more than one root at that point.

$$f(x)= (x-3)^3 (x-2)^2 (x-1)x$$

which has the derivatives

\({f}'(x)= 108+1161x^2-648x+\) \(400x^4-952x^3-84x^5+7x^6\)

\({f}''(x)=-648-2856x^2+2322x-\) \(420x^4+1600x^3+42x^5\)

\({f}'''(x)=2322+4800x^2-5712x\)+ \(210x^4-1680x^3\)

And of course we knew intuitively that \(f(x)\) has the following real roots \(3,3,3\), \(2,2\), \(1,0\) .We will study the repeated roots here which are 3 and 2.

Solving the derivatives with respect to \(x\) will yield the following:

\({f}'(x)\) has zeros when \(x\) = \(3\) , \(3\) , \(2\) , \(2.454295852\) , \(1.270906755\) , \(0.2747973935\).

\({f}''(x)\) has zeros when \(x\) = \(3\) , \(2.698648167\) , \(2.203645141\) , \(1.533922047\) , \(0.5637846449\).

\({f}'''(x)\) has zeros when \(x\) =\(2.878374203\) , \(2.445752631\) , \(1.806349702\) , \(0.8695234637\).

In General:

Let \(x_{0} \in \mathbb{R}\) be any arbitrary root for the polynomial function \(f(x)\) i.e. \(f(x_{0}) = 0\) then \(x_{0}\) said to be repeated root \(n+1\)

*times*if \(\exists m,n \in \mathbb{N}\) such that \(\forall m \leq n\) \( f^{(m)}(x_{0}) = 0 \)Exercise: Let \(p\) and \(q\) be real numbers with \(p \leq 0\). Find the co-ordinates of the turning points of the cubic \(y=x^3+px+q\). Show that the cubic equation \(x^3+px+q\) has three real roots, with two or more repeated, precisely when $$4p^3+27q^2=0$$

Under what conditions on \(p\) and \(q\) does \(x^3+px+q=0\) have (i) three distinct real roots, (ii) just one real root? How many real roots does the equation \(x^3+px+q=0\) have when \(p \geq 0\) ?

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Does this Method imply trigonometric function and other function or restricted to polynomials ?

15/12/2016 19:42 485