Homework 3 P14 1 2 3 4 SolutionWe have fx 3x2 2x 4 an

Homework 3 P14

1.   \"g(1)\" =
2.   \"g(0.1)\" =
3.   \"g(0.01)\" =
4.   \"g(0.001)\" =

Solution

We have f(x) = 3x2 + 2x + 4 and g(h) = [f (3 + h) -f ( 3 )]/h . Then f (3) = 3(3)2 + 2*3 + 4 = 27 + 6 + 4 = 37. Also f (4) = 3(4)2 + 2*4 + 4 = 48+ 8 + 4 = 60; f (3.1) = 28.83 + 6.2 + 4 = 39.03; f ( 3.01) = 27.1803 + 6.02 + 4 = 37.2003 ; f ( 3.001) = 27.018003 + 6.002 + 4 = 37.020003

Then g(1) =  [f (3 + 1) -f ( 3 )]/1 = [f (4) -f ( 3 )]/1  = 60 - 37 = 23 , g (0.1) = [f (3.1) - f ( 3 )]/0.1 = (39.03 - 37)/ 0.1 = 20.3; g ( 0.01) = [f (3.01) - f ( 3 )]/0.01 = (37.2003 - 37) / 0.01 = 20.03; g ( 0.001) = [f (3.001) - f ( 3 )]/0.001 = (  37.020003 - 37)/ 0.001 = 20.0003. These results are tabulated below:

The values of g are getting closer and closer to 20. Therefore, L = 20

1. g(1) = 23
2. g(0.1) = 20.3
3. g (0.01) = 20.03
4. g (0.001) = 20.0003
Homework 3 P14 1. = 2. = 3. = 4. = SolutionWe have f(x) = 3x2 + 2x + 4 and g(h) = [f (3 + h) -f ( 3 )]/h . Then f (3) = 3(3)2 + 2*3 + 4 = 27 + 6 + 4 = 37. Also

Get Help Now

Submit a Take Down Notice

Tutor
Tutor: Dr Jack
Most rated tutor on our site