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Concept explainers
The quadratic approximation for f(x)=x6 at a = 1.
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Answer to Problem 1PT
Quadratic approximation for f(x)=x6 at a = 1 is d) T2(x)=1+6(x−1)+15(x−1)2_.
Explanation of Solution
Result used:
The Taylor polynomial of degree n for f about a is, Tn(x)=f(a)+f′(a)1!(x−a)+f″(a)2!(x−a)2+…+fn(a)n!(x−a)n. Tn(x) may be used to approximate f(x) for any given x in the interval of convergence for the Taylor series for f about a.
Calculation:
The given function is f(x)=x6.
To find the quadratic approximation, find T2(x) where a = 1.
The Taylor polynomial of degree 2 for f(x)=x6 about a = 1 is,
T2(x)=f(1)+f′(1)1!(x−1)+f″(1)2!(x−1)2
Obtain the values of f(1),f′(1) and f″(1) as follows.
f(x)=x6;f(1)=1f′(x)=6x5;f′(1)=6f″(x)=30x4;f″(1)=30
Substitute these values in T2(x)=f(1)+f′(1)1!(x−1)+f″(1)2!(x−1)2. Then,
T2(x)=1+61(x−1)+302(x−1)2=1+6(x−1)+15(x−1)2
Therefore, the quadratic approximation is, d) T2(x)=1+6(x−1)+15(x−1)2_.
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Chapter 11 Solutions
Study Guide for Stewart's Multivariable Calculus, 8th
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