Researchers at a major car company have found a function that relates gasoline consumption to speed for a particular model of car. In particular, they have determined that the consumption, C, in liters per kilometer, at a given speed, s. is given by a function C = f(s), where s is the car's speed in kilometers per hour.

Required:
a. What are the units on f'(s)?
b. Data provided by the car company tells us that f(80) = 0.015, f(90) = 0.02, and f(100) = 0.027. Use this information to estimate the instantaneous rate of change of fuel consumption with respect to speed at s = 90

Respuesta :

Answer:

(a) The units of f'(s) is kilometer per liter

(b) [tex]\triangle = 2.22 * 10^{-4}[/tex]

Step-by-step explanation:

Given

[tex]C = f(s)[/tex]

[tex]f(80) = 0.015, f(90) = 0.02, f(100) = 0.027[/tex]

Solving (a): Unit of f'(s)

From the question, we understand that:

[tex]C = f(s)[/tex]

and

[tex]C = Litre/km[/tex] --- units

f'(s) is the inverse of C.

Hence:

i.e.

[tex]f'(s) = C^{-1}[/tex] --- units

So, we have:

[tex]f'(s) = (Litre/km)^{-1}[/tex]

[tex]f'(s) = Km/Litre[/tex]

Solving (b): Instantaneous rate of change at:

[tex]s = 90[/tex]

We have:

[tex]C = f(s)[/tex]

The change is calculated as:

[tex]\triangle = \frac{f(s)}{s}[/tex]

Substitute 90 for s

[tex]\triangle = \frac{f(90)}{90}[/tex]

Given that: [tex]f(90) = 0.02[/tex]

[tex]\triangle = \frac{0.02}{90}[/tex]

[tex]\triangle = 0.000222[/tex]

[tex]\triangle = 2.22 * 10^{-4}[/tex]