Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x − 1 = 0. (Round your answer to four decimal places.)

Answer 1

Write: d(x⁴ − x − 1)/dx equals 4x³ − 1.

Construct: x − [(x⁴ − x − 1)/(4x³ − 1)] and place x₁ = 1 in all positions of x.

Then evaluate 1 − [(1⁴ − 1 − 1)/(4 × 1³ − 1)] or 1 − -1/3 to 4/3 or 1.3333 as

the second approximation to the root of x⁴ − x − 1 = 0.

Feeding successive approximations of the root completes the chart below:

x₁ = 1 and f(x₁) = 4/3

x₂ = 4/3 and f(x₂) = 1.23580786

x₃ = 1.23580786 and f(x₃) = 1.221058994

x₄ = 1.221058994 and f(x₄) = 1.220744226

x₅ = 1.220744226 and f(x₅) = 1.220744085

x₆ = 1.220744085 and f(x₆) = 1.220744085

Note that the bottom line of the chart shows

x₆ equal to f(x₆) equal to 1.220744085.

Calculate f(x₆) or [1.220744085⁴ − 1.220744085 − 1] to obtain 2.474 × 10^-9 , extremely close to 0.