Answer:
4.6
Explanation:
There are a number of "improvement" techniques that can be used to improve an estimate of a solution to a polynomial. Some of them use the polynomial itself to form the iterator function.
Iterator
There are a number of ways we can solve the given equation for x. Often, at least one of these will converge when used for improving the estimated value of x.
x₁ = (x^3 -72)/6 . . . . . . . add 6x-72, divide by 6
x₂ = 72/(x^2 -6) . . . . . . . factor out x and divide by its coefficient
x₃ = ∛(72 +6x) . . . . . . . add 6x and take the cube root
The first two of these iterators tend to diverge. For x = 4.5, they give ...
x₁ = (4.5³ -72)/6 = 3.1875
x₂ = 72/(4.5² -6) = 5.0526
Both of these stray out of the given interval of [4, 5].
However, the third one tends to converge, so can give us a useful approximation in one iteration:
x₃ = ∛(72 +6×4.5) ≈ ∛99 ≈ 4.6261
Second iteration
Applying this again, we get ...
x ≈ ∛(72 +6·4.6261) ≈ 4.6378
The approximate root to 1 decimal place is 4.6.
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Additional comment
One of the quickest converging iterators is that provided by Newton's method. For finding the solution to f(x) = 0, it uses ...
x₁ = x -f(x)/f'(x) . . . . . . . where f'(x) is the first derivative
For this problem, we can define ...
f(x) = x^3 -6x -72 ⇒ f'(x) = 3x^2 -6
x₁ = x -(x^3 -6x -72)/(3x^2 -6) = (3x^3 -6x -x^3 +6x +72)/(3x^2 -6)
x₁ = (2x^3 +72)/(3x^2 -6)
Then for x = 4.5, the "improved" solution is ...
x₁ = (2(4.5³) +72)/(3(4.5²) -6)) = 254.25/54.75 = 339/73 ≈ 4.6438
A second iteration gives ...
x ≈ 4.639026 . . . . accurate to 5 decimal places
With this iterator, the number of accurate decimal places tends to double with each iteration.