
Andy C. answered 07/02/18
Tutor
4.9
(27)
Math/Physics Tutor
Well, you can use the SECANT line to approximate the derivatives.
Basically, you calculate the slope of the secant line between each pair of
consecutive points.
For x=0,1,2,3,4,5,6,7,8,9,10,11 representing the month number, we
are interested in approximating f'(0.5), f'(1.5), f'(2.5) etc
Since x increases by 1 along the interval, the slope of the secant
line between each pair ( 0,382.44) , (1,383.69), (2,384.24) ... etc... (11,383.69)
is just the differences between each y-value.
To estimate the derivative, we can just cut the difference in half and add it
to the previous y-value.
Specifically, f'(x+1/2) is approximated by f(x) + 1/2( f(x+1) - f(x)) =
f(x) + 1/2 f(x+1) - 1/2f(x) =
1/2 f(x+1) + 1/2 f(x) =
[f(x+1) + f(x) ]/2
which is the AVERAGE of each y-value
The results are shown in the following table
x f(x) f(x+1)-f(x) approximation = [f(x)+f(x+1)]/2
0 382.44
1 383.69 1.25 383.065
2 384.24 0.55 383.965
3 386.25 2.01 385.245
4 386.39 0.14 386.32
5 385.86 -0.53 386.125
6 384.44 -1.42 385.15
7 381.77 -2.67 383.105
8 380.74 -1.03 381.255
9 380.8 0.06 380.77
10 382.33 1.53 381.565
11 383.69 1.36 383.01
1 383.69 1.25 383.065
2 384.24 0.55 383.965
3 386.25 2.01 385.245
4 386.39 0.14 386.32
5 385.86 -0.53 386.125
6 384.44 -1.42 385.15
7 381.77 -2.67 383.105
8 380.74 -1.03 381.255
9 380.8 0.06 380.77
10 382.33 1.53 381.565
11 383.69 1.36 383.01