Question

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) = 2x − 4x3, a = −1 [infinity] f n(−1) n! (x + 1)n n = 0 = 2 − 10(x + 1) + 12(x + 1)2 − 4(x + 1)3 [infinity] f n(−1) n! (x + 1)n n = 0 = 2 − 10(x + 1) + 4(x + 1)2 − 12(x + 1)3 [infinity] f n(−1) n! (x + 1)n n = 0 = 2 + 10(x + 1) + 12(x + 1)2 + 4(x + 1)3 [infinity] f n(−1) n! (x + 1)n n = 0 = 2 − 12(x + 1) + 10(x + 1)2 − 4(x + 1)3 [infinity] f n(−1) n! (x + 1)n n = 0 = 2 + 10(x + 1) + 4(x + 1)2 + 12(x + 1)3

Answer 1

f(x) = 2 - 10 (1 + x) + 12 (1 + x)^2 - 4 (1 + x)^3

Explanation:

The general form of the series is shown in the attachment. For the purpose here, we need to evaluate f(-1), f'(-1), f''(-1) and so on.

f(x) = 2x -4x^3; f(-1) = 2(-1)(1 -2(-1)^2) = (-2)(-1) = 2

f'(x) = 2 -12x^2; f'(-1) = 2 -12(-1)^2 = -10

f''(x) = -24x; f''(-1) = -24(-1) = 24

f'''(x) = -24; f'''(-1) = -24

So, the series is ...

f(x) = 2 -10(x +1)/1! +24(x +1)^2/2! -24(x +1)^3/3!

f(x) = 2 -10(x +1) +12(x +1)^2 -4(x +1)^3 . . . . . . . matches the first choice