Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess [tex]x_{0}[/tex] for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

[tex]x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}[/tex]

For the information given:

[tex]f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262[/tex]

For the initial value [tex]x_{0}[/tex] you can choose [tex]x_{0}=0[/tex] although you can choose any value that you want.

So for approximation [tex]x_{1}[/tex]

[tex]x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021[/tex]

Next, with [tex]x_{1}=0.0496021[/tex] you put it into the equation

[tex]f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150[/tex], you can see that this value is close to 0 but we need to refine our solution.

For approximation [tex]x_{2}[/tex]

[tex]x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883[/tex]

Again we put [tex]x_{2}=0.168883[/tex] into the equation

[tex]f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307[/tex] this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation [tex]x_{5}[/tex] gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.