First sketch the graphs on each side of the equation and approximate the roots of the equation by approximating where the graphs cross
Approximate values for the roots are `x=0.2` and `1.2`
The formula for Newton's method is
`x_(n+1) = x_n - f(x_n)/(f'(x_n))`
Now, `f(x) = (x^2-x+1) +(-5e^(-x^2)sinx)`
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First sketch the graphs on each side of the equation and approximate the roots of the equation by approximating where the graphs cross
Approximate values for the roots are `x=0.2` and `1.2`
The formula for Newton's method is
`x_(n+1) = x_n - f(x_n)/(f'(x_n))`
Now, `f(x) = (x^2-x+1) +(-5e^(-x^2)sinx)`
`implies`
`f'(x) = (2x - 1 )+(10xe^(-x^2)sinx -5e^(-x^2)cosx)`
Working on the first root `x approx 0.2` gives the sequence
x0 = 0.2
x1 = 0.17677893
x2 = 0.17721437
x3 = 0.17721451
x4 = 0.17721451 `implies` root is 0.17721451 to 8dp
Working on the second root `x approx 1.2` gives the sequence
x0 = 1.2
x1 = 1.16247348
x2 = 1.16224076
x3 = 1.16224075
x4 = 1.16224075 `implies` root is 1.16224075 to 8dp
The roots are 0.17721451, 1.16224075 to 8dp