Consider the following equation. 3x^4 − 8x^3 + 9 = 0, [2, 3]

Question

(b) Use Newton's method to approximate the root correct to six decimal places. 

Answer 1

Write the statement x − (3x⁴ − 8x³ + 9) ÷ d(3x⁴ − 8x³ + 9)/dx which is equal to

x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²).

Better yet, type x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²) into a programmable calculator

and, for the given interval [2,3], evaluate x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²) for the

"middle value" of 2.5.

Take the result of this evaluation (2.468333333) and feed it back into

x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²) to obtain 2.46681613. Then run

2.46681613 in x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²) to yield 2.466812727.

Here, evaluation of x − (3x⁴ − 8x³ + 9)/(12x³ − 24x²) for x = 2.466812727

again gives 2.466812727. This is the signal that an extremely close

approximation of the root for 3x⁴ − 8x³ + 9 = 0 in [2,3] has been reached.

Running x = 2.466812727 in 3x⁴ − 8x³ + 9 will return something like

-6.8×10^-9 which is zero for all practical purposes.

Round 2.466812727 to the millionths decimal place to obtain x = 2.466813.