How do you use Newton's Method to approximate the root of the equation #x^4-2x^3+5x^2-6=0# on the interval #[1,2]# ?

Question

Luke Alvarez · Accepted Answer

The answer is #1.217562155#. Recall that Newton's Method uses the formula: #x_(n+1)=x_n−(f(x_n))/(f'(x_n))# The equation is already a function, so: #f(x)=x^4-2x^3+5x^2-6# And we need the derivative: #f'(x)=4x^3-6x^2+10x# The easiest way to iterate is to program your calculator. Enter #f(x)# into #Y_1# and #f'(x)# into #Y_2#. Then enter a very short program that does this: #A−(Y_1(A))/(Y_2(A))->A# You can go to my website for specific instructions for the TI-83 or the Casio fx-9750 . Finally, you need a starting value, #x_1#. Since the question is asking for a root in the interval #[1,2]#, we should start at #x=1.5#. So enter the following into your calculator, #1->A# Then execute the program until you get the desired accuracy: #1.2625#
#1.218807774#
#1.217563128#
#1.217562155#
#1.217562155# We get 3 digits of accuracy after 2 iterations, 6 after 3 iterations, and 10 after 4 iterations. So the answer converges very quickly for this root.

Axel Alvarez · Answer

undefined To use Newton's Method to approximate the root of the equation $x^4 - 2x^3 + 5x^2 - 6 = 0$ on the interval [1, 2], follow these steps:

1. Choose an initial guess $x_0$ within the interval [1, 2].
2. Compute the derivative of the function: $f'(x) = 4x^3 - 6x^2 + 10x$.
3. Apply the Newton's Method iteration formula: 
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
4. Repeat the iteration until the desired level of accuracy is achieved or until the number of iterations reaches a predetermined limit.

Given that the function is $f(x) = x^4 - 2x^3 + 5x^2 - 6$, and its derivative is $f'(x) = 4x^3 - 6x^2 + 10x$, you can now follow these steps to perform Newton's Method with an initial guess within the interval [1, 2].