Question

The equation f(x) is given as x2_4=0. Considering the initial approximation at

x0=6 then the value of x1 is given as

Select one:

O A. 10/3

O B. 7/3

O C. 13/3

O D. 4/3

Answer 1

The value of
`x_(1)` is given by
`(10)/(3)`. Hence, the answer is A.

Explanation:

This exercise represents a case where the Newton-Raphson method is used, whose formula is used for differentiable function of the form
`f(x) = 0`. The expression is now described:

`x_(n+1) = x_(n) - (f(x_(n)))/(f'(x_(n)))}`

Where:

`x_(n)` - Current approximation.

`x_(n+1)` - New approximation.

`f(x_(n))` - Function evaluated in current approximation.

`f'(x_(n))` - First derivative of the function evaluated in current approximation.

If
`f(x) = x^(2) - 4`, then
`f'(x) = 2\cdot x`. Now, given that
`x_(0) = 6`, the function and first derivative evaluated in
`x_(o)` are:

`f(x_(o)) = 6^(2) - 4`

`f(x_(o)) = 32`

`f'(x_(o))= 2 \cdot 6`

`f'(x_(o)) = 12`

`x_(1) = x_(o) - (f(x_(o)))/(f'(x_(o)))`

`x_(1) = 6 - (32)/(12)`

`x_(1) = 6 - (8)/(3)`

`x_(1) = (18-8)/(3)`

`x_(1) = (10)/(3)`

The value of
`x_(1)` is given by
`(10)/(3)`. Hence, the answer is A.