Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to four decimal places.) f(x) = cos x, x1 = 0.7

The two iterations of f(x) = 1.5598

Step-by-step explanation:

If we apply  Newton's iterations method, we get a new guess of a zero of a function, f(x), xₙ₊₁, using a previous guess of, xₙ.

xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)

Given;

f(xₙ) = cos x, then  f'(xₙ) = - sin x

cos x / - sin x = -cot x

substitute in "-cot x" into the equation

xₙ₊₁ = xₙ - (- cot x)

xₙ₊₁ = xₙ + cot x

x₁ = 0.7

first iteration

x₂ = 0.7 + cot (0.7)

x₂ =  0.7 + 1.18724

x₂ = 1.88724

second iteration

x₃ = 1.88724 + cot (1.88724)

x₃ = 1.88724 - 0.32744

x₃ = 1.5598

To four decimal places = 1.5598

differnt varibles efect the range

step-by-step explanation: