Sample Problem 2: Determine the absolute error and relative error when approximating the derivative of f(x) = -x² + 5x at x=2 using the approximation formula, f'(x) = f(x+h)-f(x) h with h=0.1 Solution: The approximation formula represents the approximated value while the true value is simply the derivative f'(x) at x=2. AV: f'(x) = f(x + h) − f (x) _ [-(x + 0.1)² +5(x + 0.1)] - [-x² + 5x] h 0.1 f'(x) =L-(2 +0.1)? +5(2 + 0.1)] −{-(2)? +5(2)] 0.1 f'(2) = 6.09 - 6 0.1 = 0.9 TV: f'(x) = -2x+5 f'(2) = -2(2) +5=1 Thus, ε = 10.9 = 0.1 0.1 Er == 0.1
Sample Problem 2: Determine the absolute error and relative error when approximating the derivative of f(x) = -x² + 5x at x=2 using the approximation formula, f'(x) = f(x+h)-f(x) h with h=0.1 Solution: The approximation formula represents the approximated value while the true value is simply the derivative f'(x) at x=2. AV: f'(x) = f(x + h) − f (x) _ [-(x + 0.1)² +5(x + 0.1)] - [-x² + 5x] h 0.1 f'(x) =L-(2 +0.1)? +5(2 + 0.1)] −{-(2)? +5(2)] 0.1 f'(2) = 6.09 - 6 0.1 = 0.9 TV: f'(x) = -2x+5 f'(2) = -2(2) +5=1 Thus, ε = 10.9 = 0.1 0.1 Er == 0.1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
What is the greatest value of h that can be Use the formula in approximating the derivative
(same as Sample Problem 2) of
f(x) = 2x³- 3x + 1 at x=1 using h=0.5 Compute the absolute and relative errors
Ps. The picture below is Sample Prob. 2 you can base your solution there
![Sample Problem 2:
Determine the absolute error and relative error when approximating the derivative of f(x) = -x² + 5x at
x=2 using the approximation formula,
f'(x) =
f(x+h)-f(x)
h
with h=0.1
Solution: The approximation formula represents the approximated value while the true value is simply
the derivative f'(x) at x=2.
AV:
f(x + h) = f(x) _ [−(x + 0.1)² + 5(x + 0.1)] - [-x² + 5x]
f'(x) =
h
0.1
[-(2+0.1)² + 5(2 + 0.1)] − [−(2)² +5(2)]
f'(x) =
0.1
6.09
6
f'(2) =
= 0.9
TV:
Thus,
0.1
f'(x) = -2x+5
f'(2) = -2(2) +5=1
ε = 10.9|= 0.1
0.1
Ep. = = 0.1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9fdba58-fccf-41bc-a7d5-686b96385393%2F4de69bb5-d50c-404c-b2e9-563b7cc6cedc%2F1icrn8q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Sample Problem 2:
Determine the absolute error and relative error when approximating the derivative of f(x) = -x² + 5x at
x=2 using the approximation formula,
f'(x) =
f(x+h)-f(x)
h
with h=0.1
Solution: The approximation formula represents the approximated value while the true value is simply
the derivative f'(x) at x=2.
AV:
f(x + h) = f(x) _ [−(x + 0.1)² + 5(x + 0.1)] - [-x² + 5x]
f'(x) =
h
0.1
[-(2+0.1)² + 5(2 + 0.1)] − [−(2)² +5(2)]
f'(x) =
0.1
6.09
6
f'(2) =
= 0.9
TV:
Thus,
0.1
f'(x) = -2x+5
f'(2) = -2(2) +5=1
ε = 10.9|= 0.1
0.1
Ep. = = 0.1
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