For session 2 dealing with derivatives as a rate of change. In the problem pdf for question 2. Taking f'(10) why do we automatically assume to take f(11) to find the value of 175 in order to validate the $40 dollar decrease?
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
f'(10) is the rate at which the bank balance is changing during the month October (the tenth month). f(10) is the actual balance in October and f(11) the balance in November.
By subtracting f(10) from f(11) you can simply check how much the balance has changed from October to November. In this case:
\[f(11)-f(10) = 175 - 220 = -45\]
Which is close enough to -40 to verify the rate of change found by taking f'(10).
On a side note. The solution pdf suggest trying f(10)-f(11) to check the change but that's a bad idea. The difference will still be 45 but it will be a positive number when the actual change has been a decrease. As a rule, subtract function values for smaller x's from function values for larger x's to not just get a change, but also whether that change has been positive or negative.