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
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.
1 answer
1 answer