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+ t; 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)
c(10) = 10 +4 = 14
c(2) = 10
14 -10 = 4
To determine the best days to park in the garage, we need to compare the costs for each day within the given time frame.
a) The function provided shows that the cost to park is $10 for the first 2 days (0 <= t <= 2), then increases by $1 per day for the next 3 days (2 < t <= 5), and finally drops to $7 for the last 2 days (5 < t <= 7).
Based on this information, the best days to park in the garage would be the days with the lowest costs. In this case, the best days would be the first 2 days (0 <= t <= 2), as the cost is the lowest at $10 per day.
b) To calculate c(4) - c(2), we substitute the values for t in the function.
c(4) = 10 + 4 = 14 (Substituting t = 4 into the function)
c(2) = 10 (Substituting t = 2 into the function)
Now, we can subtract c(2) from c(4) to find the difference.
c(4) - c(2) = 14 - 10 = 4
Therefore, c(4) - c(2) is equal to 4.