Consider a function 9 q (x) = (x – 3) 2 defined for x > 3. Find the expression for the remainder term r3 (x) in the Taylor series for this function centered at æo = 6. 39.375 (x–6)³ (ã–3)1.5 r3 (x) = 3! 15.75 (x–3)³ (5–3)2.5 3! r3 (z) 39.375 (2–3)³ (ž–3)1.5 r3 (z) = 3! 15.75 (7–6)³ (5–3)2.5 3! r3 (x) = 39.375 (æ–3)³ (ã–6) 1.5 (x) T3 3! 15.75 (r–3)³ (ã–6)2.5 O r3 (2) 3! Where a has a value between x and 6.
Consider a function 9 q (x) = (x – 3) 2 defined for x > 3. Find the expression for the remainder term r3 (x) in the Taylor series for this function centered at æo = 6. 39.375 (x–6)³ (ã–3)1.5 r3 (x) = 3! 15.75 (x–3)³ (5–3)2.5 3! r3 (z) 39.375 (2–3)³ (ž–3)1.5 r3 (z) = 3! 15.75 (7–6)³ (5–3)2.5 3! r3 (x) = 39.375 (æ–3)³ (ã–6) 1.5 (x) T3 3! 15.75 (r–3)³ (ã–6)2.5 O r3 (2) 3! Where a has a value between x and 6.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Calculus
Transcribed Image Text:Consider a function
9
9 («) — (ӕ — 3) 2
defined for a > 3. Find the expression for the remainder term r3 (x)
in the Taylor series for this function centered at æo = 6.
39.375 (r-6)³ (õ–3)1.5
r3 (x)
3!
15.75 (г—3)3 (а -3)2.5
(z)
3!
39.375 (r–3)³ (ž–3)1.5
r3 (x) :
3!
15.75 (x-6)³ (ã–3) 2.5
3!
r3
(z)
39.375 (z–3)³ (7–6)1.5
"3 (x) =
3!
15.75 (z–3)³ (5–6)2.5
(x) =
3!
Where a has a value between x and 6.
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