A parking garage is changing its parking fees for a week long festival. The function below represents the cost to park in the garage as a function of time (days into the festival).
Piece wise
c(t) =
10; 0 <= t <= 2
10; 2 < t <= 5
7; 5 < t <= 7
a) What would be the best days to park in the garage?
b) calculate c(4) - c(2)
a) To determine the best days to park in the garage, we need to find the days where the cost of parking is the lowest. Looking at the given piecewise function, we can see that the cost of parking is $10 for the first two days (0 <= t <= 2), drops to $7 from day 5 to day 7 (5 < t <= 7), and stays at $10 for the remaining days (2 < t <= 5). Therefore, the best days to park in the garage would be days 5, 6, and 7, when the cost is $7.
b) To calculate c(4) - c(2), we substitute the values into the given function:
c(4) = 10 (since 2 < t <= 5, and 4 falls within this range)
c(2) = 10 (since 0 <= t <= 2, and 2 falls within this range)
Now we can calculate c(4) - c(2):
c(4) - c(2) = 10 - 10 = 0
Therefore, c(4) - c(2) equals 0.