Answer:
a) x ≈ 2.794
b) x ≈ 1.9129
Explanation:
You want a root of f(x) = x² -x -5 to 3 decimal places using the bisection method starting with interval [2.7, 3] and ε = 0.05. You also want the root of f(x) = x³ -7 to 4 decimal places using Newton's method iteration starting from p0 = 3 and ε = 0.0005.
a) Quadratic
The bisection method works by reducing the interval containing the root by half at each iteration. The function is evaluated at the midpoint of the interval, and that x-value replaces the interval end with the function value of the same sign.
For example, the middle of the initial interval is (2.7+3)/2 = 2.85, and f(2.85) has the same sign as f(3). The next iteration uses the interval [2.7, 2.85].
The attached table shows that successive intervals after bisection are ...
[2.7, 3], [2.7, 2.85], [2.775, 2.85], [2.775, 2.8125], [2.775, 2.79375]
The right end of the last interval gives a value of f(x) < 0.05, so we feel comfortable claiming that as a solution to the equation f(x) = 0.
x ≈ 2.794
b) Cubic
Newton's method works by finding the x-intercept of the linear approximation of the function at the last approximation of the root. The next guess (x') is found using the formula ...
x' = x - f(x)/f'(x)
where f'(x) is the derivative of the function.
Many modern calculators can find the function derivative, so this iteration function can be used directly by a calculator to give the next approximation of the root. That is shown in the bottom of the attachment.
If you wanted to write the iteration function for use "by hand", it would be ...
x' = x -(x³ -7)/(3x²) = (2x³ +7)/(3x²)
Starting from x=3, the next "guess" is ...
x' = (2·3³ +7)/(3·3²) = 61/27 = 2.259259...
When the calculator is interactive and produces the function value as you type its argument, you can type the argument to match the function value it produces. This lets you find the iterated solution as fast as you can copy the numbers. No table is necessary.
In the attachment, the x-values used for each iteration are rounded to 4 decimal places in keeping with the solution precision requirement. The final value of x shown in the table gives ε < 0.0005, as required.
x ≈ 1.9129
__
Additional comment
The roots to full calculator precision are ...
quadratic: x ≈ 2.79128784748; exactly, 0.5+√5.25
cubic: x ≈ 1.91293118277; exactly, ∛7
The bisection method adds about 1/3 decimal place to the root with each iteration. That is, it takes on average about three iterations to improve the root by 1 decimal place.
Newton's method approximately doubles the number of good decimal places with each iteration once you get near the root. Its convergence is said to be quadratic.
<95141404393>