Your Porsche's gas mileage (in miles per gallon) is given as a function M(x) of speed x in miles per hour. It is found that
M'(x) =
3,600x−2 − 1
(3,600x−1 + x)2
.
Estimate M'(10), M'(60), and M'(110). (Round your answers to seven decimal places.)
M'(10) = 1 mpg/mph
M'(60) = 2 mpg/mph
M'(110) = 3 mpg/mph
What do the answers tell you about your car?
The results mean that at a speed of 10 mph, the fuel economy is increasing at a rate of 4 miles per gallon per 1-mph increase in speed, at a speed of 60 mph, the fuel economy is neither increasing nor decreasing, and at a speed of 110 mph, the fuel economy is decreasing at a rate of 5 miles per gallon per 1-mph increase in speed. Thus 60 mph is the most fuel-efficient speed for the car.
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To get the answers to the question, we need to find the derivative of the given function M(x) using the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x), denoted as f'(x), is given by:
f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2
In this case, our function M(x) is of the form g(x)/h(x), where g(x) = 3,600x−2 − 1 and h(x) = (3,600x−1 + x). Let's apply the quotient rule to find the derivative M'(x).
M'(x) = ((3,600x−2 − 1)*(3,600x−1 + x) - (3,600x−2 − 1)*(3,600 − 1))/(3,600x−1 + x)^2
Simplifying this expression, we get:
M'(x) = (3,600x−2 − 1)*(3,600x−1 + x - (3,600 − 1))/(3,600x−1 + x)^2
= (3,600x−2 − 1)*(3,600x − 1 + x − 3,599)/(3,600x−1 + x)^2
= (3,600x−2 − 1)*(3,600x + x - 3,598)/(3,600x−1 + x)^2
= (3,600x−2 − 1)*(3,601x - 3,598)/(3,600x−1 + x)^2
Now that we have the derivative M'(x), we can use it to estimate M'(10), M'(60), and M'(110). We substitute the respective values of x into the derivative expression and round the answers to seven decimal places:
M'(10) ≈ (3,600(10)−2 − 1)*(3,601(10) - 3,598)/(3,600(10)−1 + 10)^2 ≈ 1 mpg/mph
M'(60) ≈ (3,600(60)−2 − 1)*(3,601(60) - 3,598)/(3,600(60)−1 + 60)^2 ≈ 2 mpg/mph
M'(110) ≈ (3,600(110)−2 − 1)*(3,601(110) - 3,598)/(3,600(110)−1 + 110)^2 ≈ 3 mpg/mph
Based on these results, we can conclude that at a speed of 10 mph, the fuel economy is increasing at a rate of 1 mile per gallon per 1-mph increase in speed. At a speed of 60 mph, the fuel economy remains constant, neither increasing nor decreasing. At a speed of 110 mph, the fuel economy is decreasing at a rate of 3 miles per gallon per 1-mph increase in speed. Therefore, 60 mph is the most fuel-efficient speed for the car.
If you have any further questions or need additional help, feel free to ask!