Approximate f by a Taylor polynomial with degree n at the number a. Step 1 The Taylor polynomial with degree n = 3 is T3(x) = f(a) + f '(a)(x − a) + f ''(a) 2! (x − a)2 + f '''(a) 3! (x − a)3. The function f(x) = e2x2 has derivatives f '(x) = $$4x e2x2, f ''(x) = $$16x2+4 e2x2, and f '''(x) = $$48x+64x3 e2x2. Step 3 Therefore, T3(x) = . Submit Skip (you cannot come back)

If you center the series at x=1

Where is the error.

Step-by-step explanation:

From the information given we know that

  (This comes from the chain rule )

  (This comes from the chain rule and the product rule)

 (This comes from the chain rule and the product rule)

If you center the series at x=1  then

Where is the error.

1. credits

2. dual - aspect

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see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2