Calculus BC
Section ______
~Worksheet on Lagrange Error Bound (day 2)
Name __________________________
Date ___________________________
1. Let f be a function that has derivatives of all orders for all real numbers x Assume that
4
f  5  6, f   5  8, f   5  30, f   5  48, and f    x   75
for all x in the interval 5, 5.2 .
(a) Find the third-degree Taylor polynomial about x = 5 for f  x  .
(b) Use your answer to part (a) to estimate the value of f  5.2 . What is the maximum possible error in
making this estimate? Give three decimal places.
(c) Use your answer to (b) to find an interval [a, b] such that a  f  5.2  b . Give three decimal places.
(d) Could f  5.2 equal 8.254? Show why or why not.


2. Let f be the function given by f  x   cos  2 x 

 and let P  x  be the third-degree Taylor polynomial
6
for f about x = 0.
(a) Find P  x  .
1
1
1
.
  P  
 10 
 10  12, 000
f
(b) Use the Lagrange error bound to show that
3. (Calc) Let f be a function that has derivatives of all orders. Assume f  3  1,
f   3 
1
2
, f   3  
3
 4
, f   3   , and the graph of f  x  on [3, 4]
4
8
1
4
is shown on the right. The graph of f    x 
(a) Find the third-degree Taylor polynomial about x = 3 for the function f.
(b) Use your answer to part (a) to estimate the value of f  3.7  .
(c) Use information from the graph of y  f
 4
(4, 6)
is increasing on [3, 4].
 x
(3, 2)
to show
that f  3.7   P  3.7   0.08.
4
Graph of f    x 
(d) Could f  3.7  equal 1.283? Show why or why not.
4. (Calc) Find the maximum error incurred by approximating the sum of the series
1
2
3
4
n 1  n  1 
1



 ...    1 
  ... by the sum of the first five terms. Justify your answer.
2! 3! 4! 5!
 n! 


5. Let f be the function given by f  x   cos  3 x 

 and let P  x  be the fourth-degree Taylor polynomial
6
for f about x = 0.
(a) Find P  x  .
(b) Use the Lagrange error bound to show that
6.
Use series to find an estimate for
1  x2
0 e
1
6
1
6
f    P  
1
3000
.
dx so that the error is less than
1
200
. Justify your answer.
7. (Calc) Suppose a function f is approximated with a fourth-degree Taylor polynomial about x = 1. If the
5
maximum value of the fifth derivative between x = 1 and x = 3 is 0.01, that is, f    x   0.01 , find the
maximum error incurred using this approximation to compute f  3 .
Answers to Worksheet on Lagrange Error Bound
1. (a) 6  8  x  5   15  x  5   8  x  5 
2
3
(b) f  5.2  P3  5.2  8.264
R3  5.2   0.005
(c) 8.259  f  5.2  8.269
(d) No, f  5.2 can’t equal 8.254 because 8.254 does not lie in the interval found in part (c).
3
2 3 2 4 3
x
x  x
2
2!
3!
2. (a)
4
1
 1 
16  
24  4 4 
1
1
1
1
 10    2  5   1 


(b) R3   
4
4!
4!
5  4! 625  24 15, 000 12, 000
 10 
3  x  3
x  3  x  3


3. (a) 1 
2
4  2!
8  3!
2
(b) 1.310
(c) Since f
 4
Error 
2
 x  is increasing on [3, 4], f 4  x   6 on [3, 3.7] so
4
6  3.7  3
 0.060  0.08 .
4!
(d) Yes, 1.250  f  3.7   1.370 so f  3.7  could equal 1.283.
4. The series has terms that are alternating in sign, decreasing in magnitude, and having a limit of 0
so the error is less than the absolute value of the first truncated term by the Alternating Series
Remainder.
Error  6th term so Error 
5
or 0.012 .
6!
3 3 x 9 3 x 2 27 x3 81 3 x 4




5. (a) P  x  
2
2
2  2!
2  3!
2  4!
f 5  z  x  0 
(b) R4  x  
5!
5
5
243x5
1
1
1
 1   243   1 

so R4    


  
5
5!
120 32  3000
 6   5!   6  5!2
6. The series has terms that are alternating in sign, decreasing in magnitude, and having a limit of 0
so the error is less than the first truncated term by the Alternating Series Remainder.
1  x2
0 e
7. 0.003
dx  1 
1
1
1
43
1
1




. Error 
.
3 10 42 105
216 200