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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = 8x − 3x3, a = −2 [infinity] f n(−2) n! (x + 2)n n = 0 = 8 − 28(x + 2) + 18(x + 2)2 − 3(x + 2)3 [infinity] f n(−2) n! (x + 2)n n = 0 = 8 − 28(x + 2) + 3(x + 2)2 − 18(x + 2)3 [infinity] f n(−2) n! (x + 2)n n = 0 = 8 + 28(x + 2) + 18(x + 2)2 + 3(x + 2)3 [infinity] f n(−2) n! (x + 2)n n = 0 = 8 − 18(x + 2) + 28(x + 2)2 − 3(x + 2)3

Respuesta :

Answer:

The answer is

[tex]f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }[/tex]

Step-by-step explanation:

Remember that Taylor says that

[tex]f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }[/tex]

For this case

[tex]f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18[/tex]

[tex]f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }[/tex]

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