Answer:
a) see your problem statement for the explanation
b) 2.54539334183
Step-by-step explanation:
(b) Many graphing calculators have a derivative function that lets you define the Newton's Method iterator as a function. That iterator is ...
x' = x - f(x)/f'(x)
where x' is the next "guess" and f'(x) is the derivative of f(x). In the attached, we use g(x) instead of x' for the iterated value.
Here, our f(x) is ...
f(x) = 3x^4 -8x^3 +6
An expression for f'(x) is
f'(x) = 12x^3 -24x^2
but we don't need to know that when we use the calculator's derivative function.
When we start with x=2.545 from the point displayed on the graph, the iteration function g(x) in the attached immediately shows the next decimal digits to be 393. Thus, after 1 iteration starting with 4 significant digits, we have a result good to the desired 6 significant digits: 2.545393. (The interactive nature of this calculator means we can copy additional digits from the iterated value to g(x) until the iterated value changes no more. We have shown that the iterator output is equal to the iterator input, but we get the same output for only 7 significant digits of input.)
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<em>Alternate iterator function</em>
If we were calculating the iterated value by hand, we might want to write the iterator as a rational function in Horner form.
g(x) = x - (3x^4 -8x^3 +6)/(12x^3 -24x^2) = (9x^4 -16x^3 -6)/(12x^3 -24x^2)
g(x) = ((9x -16)x^3 -6)/((12x -24)x^2) . . . . iterator suitable for hand calculation
