Question

Determine the root of the equation

f(x) = e2x- x-6 closet to the point 0.95
Leave
your answer to five decimal places

Answer 1

0.97087

Explanation:

Given the function:

`f(x) = e^(2x) - x - 6`

Its first derivative is:

`f'(x) = 2e^(2x) - 1`

Using Newton's method:

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

and starting with the point:

`x_1 = 0.95`

we get:

`f(x_1) = e^(2(0.95)) - 0.95 - 6 = -0.26410`

`f'(x_1) = 2e^(2(0.95)) - 1 = 12.37178`

`x_2 = x_1 - (f(x_1))/(f'(x_1) = 0.95 - (-0.26410)/(12.37178) = 0.97134`

`f(x_2) = e^(2(0.97134))- 0.97134 - 6 = 0.00608`

`f'(x_2) = 2e^(2(0.97134)) - 1 = 12.95485`

`x_3 = 0.97134 - (0.00608)/(12.95485) = 0.97087`

`f(x_3) = e^(2(0.97087)) - 0.97087 - 6 = -2.6240413 \cdot 10^(-7)`