Question

Question #eda23

Answer 1

`x=2.16450216` to 8dp.

Explanation:

For the second Part (Newton's Method):

We are trying to find `(0.462)^-1`

Let `alpha=(0.462)^-1`

`=> alpha= 1/0.462`
`:. 0.462alpha=1`
`:. 0.462alpha-1 = 0`

Let `f(x) = 0.462x-1` Then our aim is to solve `f(x)=0`.

First let us look at the graphs:
graph{0.462x-1 [-1.5, 3, -2, 2]}

We can see there is one solution in the interval `-2 le x le 0`.

We can find the solution numerically, using Newton-Rhapson method

`\ \ \ \ \ \ \f(x) = 0.462x-1`
`:. f'(x) = 0.462`

The Newton-Rhapson method uses the following iterative sequence

`{ (x_1,=2), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Then using excel working to 8dp we can tabulate the iterations as follows:

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is `x=2.16450216` to 8dp.

Note - Compare with result from calculator `2.164502165`