Answer:
a = 9.849
b = 20.25
c = 491.03
Step-by-step explanation:
By using Pythagoras theorem in the right triangle BDC,
(Hypotenuse)² = (Leg 1)² + (Leg 2)²
BC² = BD² + DC²
a² = 9² + 4²
a =
a =
a = 9.8489
a ≈ 9.849 units
By mean proportional theorem,
AD × DC = BD²
b × 4 = 9²
b =
b = 20.25 units
BY Pythagoras theorem in ΔADB,
AB² = AD² + BD²
c² = b² + 9²
c² = (20.25)² + 9²
c² = 410.0625 + 81
c = 491.0625
c = 491. 063 units
(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)
(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...
... x ≈ 3.0473167
(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...
... x ≈ 3.2291234
(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...
... x ≈ 3.0477377
(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...
... x ≈ 3.0466805180
_____
<em>Comment on these methods</em>
Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.
The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.
We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).
