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Matching functions with polynomials Match the following

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 67E Chapter 9.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 67E

Matching functions with polynomials Match the following six functions with the given six Taylor polynomials of order 2. Give reasons for your choices.

a.

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

b.

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

c. e2x

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

d.

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

e.

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

f. e−2x

A. p2(x)= 1+ 2x + 2x2

B. p2(x) = 1 − 6x + 24x2

C.

D. p2(x) = 1 − 2x + 4x2

E.

F. p2(x)= 1 − 2x + 2x2

Step-by-Step Solution:

Solution 67E

Step 1:

Here we have to match the functions with corresponding Taylor polynomial

We know the Taylor series of the function f(x) at a is defined as

f(x)=f(a) ++....

Maclaurin series of function f(x) is  Taylor series of the function f(x) at a=0

f(x)=f(a) ++....

 (a)

The Taylor series of the function  with centre a=0  is option (C)

That is

Since by definition of Taylor series  with centre a=0  is given by

               

                 

   Therefore

 (b)

The Taylor series of the function with centre a=0  is option (E)

That is

Since by definition of Taylor series with centre a=0  is given by

               

                 

  Therefore

 (c)

The Taylor series of the function with centre a=0  is option (A) 

That is

Since by definition of Taylor series with centre a=0  is given by

               

                 

 Therefore 

 (d)

The Taylor series of the function with centre a=0  is option (D) 

That is

Since by definition of Taylor series with centre a=0  is given by

               

                 

 Therefore 

 (e)

The Taylor series of the function with centre a=0  is option (B) 

That is

Since by definition of Taylor series with centre a=0  is given by

               

             

Therefore   

 (f)

The Taylor series of the function with centre a=0  is option (F) 

That is

Since by definition of Taylor series with centre a=0  is given by

               

                 

 Therefore 

 

 

Step 2 of 1

Chapter 9.1, Problem 67E is Solved
Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

This full solution covers the following key subjects: functions, polynomials, matching, given, match. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 67E from chapter: 9.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The answer to “Matching functions with polynomials Match the following six functions with the given six Taylor polynomials of order 2. Give reasons for your choices.a. A. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2b. A. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2c. e2xA. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2d. A. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2e. A. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2f. e?2xA. p2(x)= 1+ 2x + 2x2B. p2(x) = 1 ? 6x + 24x2C. D. p2(x) = 1 ? 2x + 4x2E. F. p2(x)= 1 ? 2x + 2x2” is broken down into a number of easy to follow steps, and 191 words. Since the solution to 67E from 9.1 chapter was answered, more than 301 students have viewed the full step-by-step answer.

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