Final answer:
The Taylor series expansion for f(x) = x³ about x = -1 is Σn=0∞ cₙ(x + 3)ⁿ. The first five coefficients are c₀ = -1, c₁ = 3, c₂ = -6, c₃ = 6, and c₄ = 0.
Step-by-step explanation:
The Taylor series expansion for the function f(x) = x³ about the point x = -1 can be written as:
Σn=0∞ cₙ(x + 3)ⁿ
To find the first five coefficients, we need to determine the values of c₀, c₁, c₂, c₃, and c₄.
To find c₀, we substitute x = -1 into the function: f(-1) = (-1)³ = -1.
Therefore, the first coefficient c₀ = -1.
To find c₁, we take the derivative of f(x) and evaluate it at x = -1: f'(x) = 3x², f'(-1) = 3(-1)² = 3.
Therefore, the second coefficient c₁ = 3.
To find c₂, we take the second derivative of f(x) and evaluate it at x = -1: f''(x) = 6x, f''(-1) = 6(-1) = -6.
Therefore, the third coefficient c₂ = -6.
To find c₃, we take the third derivative of f(x) and evaluate it at x = -1: f'''(x) = 6, f'''(-1) = 6.
Therefore, the fourth coefficient c₃ = 6.
To find c₄, we take the fourth derivative of f(x) and evaluate it at x = -1: f''''(x) = 0, f''''(-1) = 0.
Therefore, the fifth coefficient c₄ = 0.