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.