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.