Interpreting Derivatives

Assume g(v) is the fuel efficiency, in miles per gallon, of a car going at a speed of v miles per hour.
   1. What are the units of g′(v)=dg/dv?
   2. What is the practical meaning of g′(55)=−0.54?
  1. When the car is going 55 mph, the rate of change of the fuel efficiency decreases to approximately 0.54 miles/gal.
  2. When the car is going 55 mph, the rate of change of the fuel efficiency decreases by approximately 0.54 miles/gal.
  3. If the car speeds up from 55 mph to 56 mph, then the fuel efficiency is approximately −0.54 miles per gallon.
  4. If the car speeds up from 55 mph to 56 mph, then the car becomes less fuel efficient by approximately 0.54 miles per gallon.
Let A = f(t) be the depth of tread, in centimeters, on a radial tire as a function of the time elapsed t, in months, since the purchase of the tire.
   3. Interpret the following in practical terms, paying close attention to units.
  1. f(6)=0.5
  2. f −1(0.31)=15
  3. f′(12)=−0.015
  4. (f −1)′(0.4) =−60
HINT: You may find it helpful to rewrite the last two equations in Leibniz notation (that is, with "d").