Help with this calculus problem please

You're using Newton's method to find the roots to

f(x) = g(x)   →   x ⁶ = cos(x)

Recast this as finding the roots to

h(x) = f(x) - g(x) = x ⁶ - cos(x)

First, get the tangent line approximation L(x) to h(x) at an arbitrary starting point x = x₀ ; in a small neighborhood of x₀, we have

h(x) ≈ L(x) = h(x₀) + h'(x₀) (x - x₀)

Find the root to the tangent line function:

0 = h(x₀) + h'(x₀) (x - x₀)   →   x = x₁ = x₀ - h(x₀)/h'(x₀)

x₁ is the first approximation of a root to h(x). Repeat the process - that is, take the tangent line to h(x) at x₁ to find the next approximation x₂, and so on - until you meet the required condition.

For the smaller root, take x₀ = -π/2 :

x₀ ≈ -1.5708

x₁ = x₀ - h(x₀)/h'(x₀) ≈ -1.31348

x₂ = x₁ - h(x₁)/h'(x₁) ≈ -1.11366

x₃ = x₂ - h(x₂)/h'(x₂) ≈ -0.982446

x₄ = x₃ - h(x₃)/h'(x₃) ≈ -0.928013

x₅ = x₄ - h(x₄)/h'(x₄) ≈ -0.920039

x₆ = x₅ - h(x₅)/h'(x₅) ≈ -0.919887

Stop at x₆ because |x₆ - x₅| ≤ 0.001.

For the larger root, you can start the process over with a different seed for x₀, such as π/2. But since h(x) is an even function and symmetric about the y-axis, the larger root will just be the same root with positive sign, 0.919887.

yes

\

step-by-step explanation:

mk./km: lm

step-by-step explanation:

,mlmlm

; mk

lk