The Taylor expansion of f for x→ 10 is f(x) = 10 + 3(x-10)² + 2(x-10)³ + 0((x a) ft (10) = -12 6) fr (10) = 12 c) f(0) = 12 f" (10)=2 e) f(0) = -12 how I solved f(x) = 6(x-10) + 6(x-10)² f"(x) = 6+12(2-10) f(2)=12 is it bor c ???
The Taylor expansion of f for x→ 10 is f(x) = 10 + 3(x-10)² + 2(x-10)³ + 0((x a) ft (10) = -12 6) fr (10) = 12 c) f(0) = 12 f" (10)=2 e) f(0) = -12 how I solved f(x) = 6(x-10) + 6(x-10)² f"(x) = 6+12(2-10) f(2)=12 is it bor c ???
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 37E
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Transcribed Image Text:The Taylor expansion of f for x→ 10 is f(x) = 10+ 3(x-10)² + 2(x-10)²³ + o((x-101³) then:
a) f(10)=-12
b) f" (10) = 12
c) f" (0) = 12
d) f" (10) = 2
e) f (0)= -12
how I solved
f'(x) = 6(x-10) + 6(x-10) ²
f(x)=6+12(x-10)
f(2)=12 is it bor c ???
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