Expand in powers of (x+2) using Taylor’s theorem.
Answers
Answer:
Step-by-step explanation:
Given : 3x³ - 2x²+ x + 4
To Find : Expand in powers of (x+2) using Taylors theorem
Solution:
f(x) = 3x³ - 2x²+ x + 4
f"(x) = 9x² - 4x + 1
f''(x) = 18x - 4
f'''(x) = 18
Using Taylors theorem
f(x) = f(c) + f'(c)(x - c) + f''(c)(x - c)² / 2! + f'''(c)(x - c)³/3! as after that all terms are zero in this case
taking c = - 2
=> f(x) = f(-2) + f'(-2)(x +2) + f''(-2)(x+2)² / 2! + f'''(-2)(x +2)³/3!
f(-2) = 3(-2)³ - 2(-2)²+ (-2) + 4 = - 30
f"(-2) = 9(-2)² - 4(-2) + 1 = 45
f''(-2) = 18(-2) - 4 = - 40
f'''(-2) = 18
=> f(x) = -30 + 45(x +2) - 40(x+2)² / 2! + 18(x +2)³/3!
=> f(x) = -30 + 45(x +2) - 20(x+2)² + 3(x +2)³
=> f(x) = 3(x +2)³ - 20(x+2)² + 45(x +2) - 30
3x³ - 2x²+ x - 4 = 3(x +2)³ - 20(x+2)² + 45(x +2) - 30
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