Problem

(a) Show that the function defined by $ f(x)…

05:47

Problem 83 Hard Difficulty

Prove Taylor's Inequality for $ n = 2, $ that is, prove that if $ \mid f''' (x) \mid \le M \text { for }
\mid x - a \mid \le d, $ then

$ \mid R_2 (x) \mid \le \frac {M}{6} \mid x - a \mid^3 $ for $ \mid x - a \mid \le d $

Answer

$\left|R_{2}\left(x_{2}\right)\right| \leq \frac{1}{6} M|x-a|^{3}$

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 10

Taylor and Maclaurin Series

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 11

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Video Transcript

to prove the problem is proof Taylor's inequality for in musical tube. That is true that if there are derivative ABS, value is less than I am for absolutely. Of Oxman's A is less than the then at two x soiree of tracks is less than m over six times as value of Xmas eight to the power of three. Mm. As word of axeman is a is last D. So from here behalf ex prime from from X This is the last time, um, for X is greater than a and the last time they plus the than behalf integral from a two x What exit is spirit between A and a plastic us from from from two. He this last time integral from eight X Um, mhm. So on behalf years, this is of Hank's from from minus half A from prom is less than X months a and, um then we have integral of have x cramp from from h x r b u T t his last integral from a two x have from from okay class. Um, he minus a it's so we have at the tax prom minus have a problem is less than from? From a hams X minus. Mm. Class, Um, over two times X minus. Hey, square on behalf integral of I ftp from from eight x did He is less than 18 year old from a two ex prime a as I have from from eight times three months. Eight plus I'm over two times. He might stay square tiki than behalf have x minus f a. His life must, uh, have a prime times x minus a us f a from from 8/2 times Xmas a square us mm over six times X minus A to the power of three. Since our two X is equal to our fax minus f a minus, have a prom xmas a plus minus five apron, prime over two times X space square sofa. Half our two acts his last time I'm over six x minus A skill Uh huh. X is greater than a and less than plastic. A similar argument using us from from from is greater than negative arm. When X is between negative 50 plus eight and a we can prove r two X is greater than negative of three factorial times. Thanks. One stay. Q. So we have from these two results and the value of our two this last time. M over six times, absolutely of X minus Space cube.

Problem

(a) Show that the function defined by $ f(x)…

Problem 83 Hard Difficulty

Prove Taylor's Inequality for $ n = 2, $ that is, prove that if $ \mid f''' (x) \mid \le M \text { for }
\mid x - a \mid \le d, $ then

$ \mid R_2 (x) \mid \le \frac {M}{6} \mid x - a \mid^3 $ for $ \mid x - a \mid \le d $

Answer

$\left|R_{2}\left(x_{2}\right)\right| \leq \frac{1}{6} M|x-a|^{3}$

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Problem

(a) Show that the function defined by $ f(x)…

Problem 83 Hard Difficulty

Prove Taylor's Inequality for $ n = 2, $ that is, prove that if $ \mid f''' (x) \mid \le M \text { for }\mid x - a \mid \le d, $ then $ \mid R_2 (x) \mid \le \frac {M}{6} \mid x - a \mid^3 $ for $ \mid x - a \mid \le d $

Answer

$\left|R_{2}\left(x_{2}\right)\right| \leq \frac{1}{6} M|x-a|^{3}$

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Prove Taylor's Inequality for $ n = 2, $ that is, prove that if $ \mid f''' (x) \mid \le M \text { for }
\mid x - a \mid \le d, $ then

$ \mid R_2 (x) \mid \le \frac {M}{6} \mid x - a \mid^3 $ for $ \mid x - a \mid \le d $