According to the cubic formula, the zeros of the given polynomial functions are −0.37,1.59 and 2.52
Given the polynomail function
4x^3 - 15x ^2 + 10x + 6 = 0
There are three zeros of the given polynomial function since the highest degree is 3
4x^3 - 15x ^2 + 10x + 6 = 0
The only way to find the solution is to use the factorization calculator. According to the cubic formula, the zeros of the given polynomial functions are −0.37,1.59 and 2.52
Learn more on polynomial functions here: https://brainly.com/question/2833285
Answer:
[tex]x=3.099566703 \textsf{ (9 dp)}[/tex]
Step-by-step explanation:
Newton-Rhapson method
This iterative method finds roots of equations in the form f(x) = 0 when an equation cannot be solved using the usual analytical methods
This method works by finding the tangent to a function at a point [tex]x_0[/tex], and using its x-intercept for the next iteration, [tex]x_1[/tex]. Repeating the process iteratively gets you closer to the root.
Newton-Rhapson formula
[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/tex]
Given function:
[tex]f(x)=4x^3-15x^2+10x-6=0[/tex]
First check that the function is in the form f(x) = 0 ← yes!
(if it is not, rearrange it so that it is equal to zero)
Now differentiate the function:
[tex]f'(x)=12x^2-30x+10[/tex]
Substitute the function and its derivative into the N-R formula:
[tex]x_{n+1}=x_n-\dfrac{4{x_n}^3-15{x_n}^2+10{x_n}-6}{12{x_n}^2-30{x_n}+10}[/tex]
To determine which value to set as [tex]x_0[/tex] examine the function.
[tex]f(x)=4x^3-15x^2+10x-6[/tex] is a cubic function with a positive leading coefficient.
Therefore, its endpoint behavior is:
[tex]f(x) \rightarrow \infty \textsf{ as } x \rightarrow \infty\\f(x) \rightarrow -\infty \textsf{ as } x \rightarrow -\infty[/tex]
Its y-intercept is (0, -6)
It's turning points are when [tex]f'(x)=0[/tex]
Therefore, using the quadratic formula, its turning points are:
[tex]x=\dfrac{-(-30)\pm\sqrt{(-30)^2-4(12)(10)}}{2(12)}=\dfrac{15\pm\sqrt{105}}{12}[/tex]
Sketching the graph (see attached) suggests that the x-intercept will be when:
[tex]x > \dfrac{15+\sqrt{105}}{12}[/tex]
[tex]\implies x > 2.1039...[/tex]
Let's start with inputting values of x = 3 and x = 4 into the function:
[tex]f(3)=4(3)^3-15(3)^2+10(3)-6=-3\\f(4)=4(4)^3-15(4)^2+10(4)-6=50[/tex]
The x-intercept will be between values where there is a change in sign. As there is a change in sign between f(3) and f(4), the root is [tex]3 < x < 4[/tex]
Therefore, let [tex]x_0=3[/tex]
[tex]\implies x_{1}=x_0-\dfrac{4{x_0}^3-15{x_0}^2+10{x_0}-6}{12{x_0}^2-30{x_0}+10}[/tex]
[tex]\implies x_{1}=3-\dfrac{4(3)^3-15(3)^2+10(3)-6}{12(3)^2-30(3)+10}=\dfrac{87}{28}[/tex]
[tex]\implies x_{2}=x_1-\dfrac{4{x_1}^3-15{x_1}^2+10{x_1}-6}{12{x_1}^2-30{x_1}+10}=3.099605842...[/tex]
[tex]\implies x_{3}=x_2-\dfrac{4{x_2}^3-15{x_2}^2+10{x_2}-6}{12{x_2}^2-30{x_2}+10}=3.099566704...[/tex]
[tex]\implies x_{4}=x_3-\dfrac{4{x_3}^3-15{x_3}^2+10{x_3}-6}{12{x_3}^2-30{x_3}+10}=3.099566703...[/tex]
[tex]\implies x_{5}=x_4-\dfrac{4{x_4}^3-15{x_4}^2+10{x_4}-6}{12{x_4}^2-30{x_4}+10}=3.099566703...[/tex]
Therefore, [tex]x=3.099566703 \textsf{ (9 dp)}[/tex]
