10. For f(x)=lnx, construct tables, rounded to four decimals, near x=1, x=2, x=5, and x=10. Use tables to estimate f'(1), f'(2), f'(5) and f'(10). Then guess a general formula for f'(x).
I don't understand how to do number 11
11. Estimate f'(2) using the values of f in the table.
ln(0.9) = -0.1054
ln(1.0) = 0
ln(1.1) = 0.0953
(.0953 + 0.1054)/0.2 = 1.0035
ln(1.9) = 0.6419
ln(2.0) = 0.6931
ln(2.1) = 0.7419
(0.7419 - 0.6419)/0.2 = 0.5000
ln(4.9) = 1.5892
ln(5.0) = 1.6094
ln(5.1) = 1.6292
(1.6292 - 1.5892)/0.2 = 0.2000
ln(9.9) = 2.2925
ln(10.0) = 2.3026
ln(10.1) = 2.3125
(2.3125 - 2.2925)/0.2 = 0.1000
So, it appears that if f(x) = ln(x) then f'(x) = 1/x
Well, using our approximation formula, f'(2) = 1/2 = 0.5
To estimate f'(2) using the values of f in the table, we need to use the concept of finite differences.
Finite differences involve taking the difference between consecutive values in the table. In this case, we will be working with the differences between the y-values (f(x)).
Let's start by constructing the table for x = 1, 2, 5, and 10 and their respective f(x) values using the given function f(x) = lnx. We will round the values to four decimal places:
________________________
|x | f(x) |
|______|_________________|
|1 | 0.0000 |
|2 | 0.6931 |
|5 | 1.6094 |
|10 | 2.3026 |
|______|_________________|
Now, to estimate f'(2), we will calculate the first-order forward difference between the points (2, 0.6931) and (1, 0.0000).
The first-order forward difference can be calculated as follows:
Forward difference = f(x + h) - f(x)
In this case, h = 1 (since we are working with adjacent x-values), so:
Forward difference = f(2 + 1) - f(2)
= f(3) - f(2)
Substituting the values from the table:
Forward difference = f(3) - f(2)
= ln(3) - ln(2)
≈ 1.0986 - 0.6931
≈ 0.4055
Therefore, the estimate for f'(2) is approximately 0.4055.
Note: The difference between f'(2) estimated using finite differences and the actual value, which can be found using calculus, may exist due to the approximation involved in this method. The more points we consider and the smaller the interval (h) between them, the more accurate the estimation will be.
If you have any further questions or need additional clarification, please let me know!