Taylor Series Test

Taylor Series Test

12th Grade

11 Qs

quiz-placeholder

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Taylor Series Test

Taylor Series Test

Assessment

Quiz

Mathematics

12th Grade

Hard

Created by

Andrea Gassmann

Used 52+ times

FREE Resource

11 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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1.  Find the radius of convergence of the power series

  14\frac{1}{4}  

 12\frac{1}{2}  

2

0

 \infty  

2.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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What are all the values of x for which the series converges?

112<x<12-\frac{11}{2}<x<\frac{1}{2}

112x12-\frac{11}{2}\le x\le\frac{1}{2}

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(, )\left(-\infty,\ \infty\right)

112x<12-\frac{11}{2}\le x<\frac{1}{2}

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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This power series converges conditionally at x = 7. Which of the following statements about the convergence of the series at x = -2 is true ?

The series converges absolutely at x = -2.

The series converges conditionally at x =-2.

The series diverges at x =-2.

There is not enough information to determine the convergence of the series at x = -2.

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

 1+6(x1)24(x1)2+80(x1)3.-1+6\left(x-1\right)-24\left(x-1\right)^2+80\left(x-1\right)^3.  This is a third-degree Taylor polynomial for a function f about        x = 1. What is the value of    f(1)f'''\left(1\right)   ?

 403\frac{40}{3}  

 803\frac{80}{3}  

80

240

480

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The third order Taylor polynomial about x = 0 of ln(1-2x) is

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Media Image
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6.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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 If a fifth degree Taylor Polynomial centered at x =2  is used to approximate f(4), and  f(4)T5(4)B\left|f\left(4\right)-T_5\left(4\right)\right|\le B  , which of the following could be B if the graph shown is a portion of the graph of  f(6)(x).f^{\left(6\right)}\left(x\right).  .  

 83\frac{8}{3}  

 163\frac{16}{3}  

4.8

1.6

1.956

7.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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What are the first three terms of the series.

x52x76+x910+C\frac{x^5}{2}-\frac{x^7}{6}+\frac{x^9}{10}+C

x522x73+8x95+C\frac{x^5}{2}-\frac{2x^7}{3}+\frac{8x^9}{5}+C

2x558x721+4x95+C\frac{2x^5}{5}-\frac{8x^7}{21}+\frac{4x^9}{5}+C

2x552x721+2x945+C\frac{2x^5}{5}-\frac{2x^7}{21}+\frac{2x^9}{45}+C

2x55x63+8x721+C\frac{2x^5}{5}-\frac{x^6}{3}+\frac{8x^7}{21}+C

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