A=(26.5)/(x^(0.45))*+
Compute (dA)/(dx) for x=0.21. (Round your answer to two decimal places.)
| mph per stop per mile
Compute (dA)/(dx) for x=4. (Round your answer to two decimal places.)
] mph per stop per mile
How is the rate of change with respect to x of the average speed of your trip affected by the number of stops per mile?
Effect of Stopping on Average Speed According to data from a General Motors study, the average speed of your trip A (in miles per hour is related to the number of stops per mile you make on the trip x by the equation
26.5 A 0.45
dA Compute - for x = 0.21. (Round your answer to two decimal places.) dx
mph per stop per mile
dA Compute - for x 4. (Round your answer to two decimal places.) xp mph per stop per mile
How is the rate of change with respect to x of the average speed of your trip affected by the number of stops per mile?
When the derivative of A at a value of x is negative it indicates that A will [--Select--- v if the input increases by a small amount. In this case, if you make 4 stops, your average speed will [--Select--- v] if you make an additional stop.
