Find the Taylor series for f(x) centered at c.

Question

f(x) = 1/3x-2 , c = -6

Answer 1

Given f(x) = 1/(3x-2) with a center point (x₀) = -6 we first start to compute some of the derivatives of the function and look for a pattern...

Taylor Theorem states --> f(x) = sum [ f^k(x₀)/k! * (x - x₀)^k ]

f(-6) = f⁰(-6) = 1/16

f¹(x) = f'(x) = -1 * [ 1/(3x-2)² ] * 3 = -3 * [ 1/(3x-2)² ] = (-3)¹ 1! * [ 1/(3x-2)² ]

f¹(-6) = -3/16²

f²(x) = f''(x) = -1 * -2 * [ 1/(3x-2)³ ] * 3 * 3 = 18 * [ 1/(3x-2)³ ] = (-3)² 2! * [ 1/(3x-2)³ ]

f²(-6) = 18/16³

f³(x) =f'''(x) = -1 * -2 * -3 * [ 1/(3x-2)⁴ ] * 3 * 3 * 3 = -6 * 27 * [ 1/(3x-2)⁴ ] = (-3)³ 3! * [ 1/(3x-2)⁴ ]

We now see the pattern and can write the full representation for the Taylor series...

f(x) = sum_[lb-->k=0]_[ub-->k=infinity] (-3)^k k! * [ 1/(16)^(k+1) ] / k! * (x + 6)^k = (-3)^k * [ 1/(16)^(k+1) ] * (x + 6)^k

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