let f(v) be the gas consumption (in liters/km) of a car going at velocity v (in km/hour). In other words, f(v) tells you how many liter of gas the car uses to go one kilometer if it is travelling at v kilometer per hour. In addition, suppose that f(80) = 0.05 and f' (80) = 0.0004. a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) and g(v)? Hence find g(80) and g' (80). b) Let h(v) be the gas consumption in liters per hour of a car going at velocity v. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the algebraic relationship between h(v) and f(v)? Hence find h(80) and h'(80). c) How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

a) The function f(v) is consumption in liters/km and g(v) is efficiency in km/liters. So, `g(v)=1/(f(v)).`

`g(80)=1/(f(80))=1/0.05=20` km/liter

`g'(80)=-(f'(80))/(f(80))^2=-0.0004/(0.05)^2=-0.16` km/liter per km/hr.

b) Since, f(v) is consumption in liters/km and v is velocity in km/h, so

`h(v)=vf(v)`

`h(80)=80*f(80)=80*0.05=4` liters/hr

`h'(80)=f(80)+80*f'(80)=0.05+80*0.0004=0.082` liters/hr

per km/hr.

c) In simpler terms, this means that...

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a) The function f(v) is consumption in liters/km and g(v) is efficiency in km/liters. So, `g(v)=1/(f(v)).`

`g(80)=1/(f(80))=1/0.05=20` km/liter

`g'(80)=-(f'(80))/(f(80))^2=-0.0004/(0.05)^2=-0.16` km/liter per km/hr.

b) Since, f(v) is consumption in liters/km and v is velocity in km/h, so

`h(v)=vf(v)`

`h(80)=80*f(80)=80*0.05=4` liters/hr

`h'(80)=f(80)+80*f'(80)=0.05+80*0.0004=0.082` liters/hr

per km/hr.

c) In simpler terms, this means that when the speed of the car increases from 80 to 81 km/hr, the fuel efficiency decreases by about 0.16 km/liter.

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