Use the maximum magnitude of the remainder term to find the maximum error in the following approximation on the given interval.
ln left parenthesis 1 minus x right parenthesisalmost equalsnegative x minus StartFraction x squared Over 2 EndFraction
​; ​[negative 0.02​,0.02​]
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Part 1
Find the maximum error. Select the correct choice below and fill in the answer box to complete your choice.
​(Use scientific notation. Use the multiplication symbol in the math palette as needed. Do not round until the final answer. Then round to two decimal places as​ needed.) 
A.
Errorless than or equals
  
enter your response here for Mequals1
B.
Errorless than or equals
  
enter your response here for Mequals0
C.
Errorless than or equals
  
enter your response here for Mequals2
D.
Errorless than or equals
  
enter your response here for Mequals3




Use the method used in this example:
Use the maximum magnitude of the remainder term to find the maximum error in the following approximation on the given interval.
ln left parenthesis 1 plus x right parenthesisalmost equalsx minus StartFraction x squared Over 2 EndFraction
​; ​[negative 0.7​,0.7​]
Question content area bottom
Part 1
Let p Subscript n Baseline left parenthesis x right parenthesis be the Taylor polynomial of order n for​ f, where n is a fixed positive integer. Suppose there exists a number M such that StartAbsoluteValue f Superscript left parenthesis n plus 1 right parenthesis Baseline left parenthesis c right parenthesis EndAbsoluteValue less than or equals Upper M for all c between a and x inclusive. The remainder in the​ nth-order Taylor polynomial for f centered at a satisfies the inequality below.
StartAbsoluteValue Upper R Subscript n Baseline left parenthesis x right parenthesis EndAbsoluteValue less than or equals Upper M StartFraction StartAbsoluteValue x minus a EndAbsoluteValue Superscript n plus 1 Over left parenthesis n plus 1 right parenthesis exclamation mark EndFraction
Part 2
The given approximation is p 2 left parenthesis x right parenthesisequalsx minus StartFraction x squared Over 2 EndFraction
 with aequals​0, so nequals2. Differentiate to find f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis x right parenthesis.
f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis x right parenthesisequalsStartFraction 2 Over left parenthesis 1 plus x right parenthesis cubed EndFraction
Part 3
​Therefore, f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis c right parenthesis equals StartFraction 2 Over left parenthesis 1 plus c right parenthesis cubed EndFraction
​, where c is between 0 and x on the interval ​[negative 0.7​,0.7​]. Now determine which value of c on the interval ​[negative 0.7​,0.7​] will maximize the magnitude of f 3 left parenthesis c right parenthesis. Notice that the smaller the value of​ c, the smaller the denominator in StartAbsoluteValue f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis c right parenthesis EndAbsoluteValue. ​Therefore, StartAbsoluteValue f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis c right parenthesis EndAbsoluteValue is the largest when cequalsnegative 0.7.
Part 4
Evaluate StartAbsoluteValue f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis c right parenthesis EndAbsoluteValue at cequalsnegative 0.7​, rounding to two decimal places.
StartAbsoluteValue f Superscript left parenthesis 3 right parenthesis Baseline left parenthesis negative 0.7 right parenthesis EndAbsoluteValueequals74.07
Part 5
​Therefore, we can use Mequals75. ​Similarly, the factor StartAbsoluteValue x cubed EndAbsoluteValue has its maximum magnitude at x equals plus or minus 0.7. Since we used cequalsnegative 0.7​, use xequalsnegative 0.7.
Part 6
Now use StartAbsoluteValue Upper R Subscript n Baseline left parenthesis x right parenthesis EndAbsoluteValue less than or equals Upper M StartFraction StartAbsoluteValue x minus a EndAbsoluteValue Superscript n plus 1 Over left parenthesis n plus 1 right parenthesis exclamation mark EndFraction
 with nequals​2, aequals​0, Mequals75​, and xequalsnegative 0.7 to find the maximum error in approximating​ f(x) by p 2 left parenthesis x right parenthesis​, rounding to two decimal places.
StartAbsoluteValue Upper R 2 left parenthesis x right parenthesis EndAbsoluteValue
equals
left parenthesis 75 right parenthesis StartFraction StartAbsoluteValue negative 0.7 minus 0 EndAbsoluteValue Superscript 2 plus 1 Over left parenthesis 2 plus 1 right parenthesis exclamation mark EndFraction
StartAbsoluteValue Upper R 2 left parenthesis negative 0.7 right parenthesis EndAbsoluteValue
equals
4.29
Simplify.
Part 7
​Therefore, the maximum error in approximating​ f(x) by p 2 left parenthesis x right parenthesis does not exceed 4.29.