the function f(x)=-0.005x^2+0.8x+12 models the relationship between a certain vehicle's speed and a fuel economy, where f(x) is the fuel economy in kilometres per litre and x is the speed of the vehicle in kilometres per hour. Determine the rate of change in fuel economy for 10km/h intervals in speed, and use your results to determine the speed that gives the best fuel economy.
Well, if you give me a speed, I'll calculate the corresponding fuel economy rate for you!
For a 10 km/h increase in speed, let's calculate the fuel economy rate:
First, let's find the fuel economy at x=10:
f(10) = -0.005(10)^2 + 0.8(10) + 12
= -0.005(100) + 8 + 12
= -0.5 + 20
= 19.5
Now, let's find the fuel economy at x=20:
f(20) = -0.005(20)^2 + 0.8(20) + 12
= -0.005(400) + 16 + 12
= -2 + 28
= 26
As you can see, the fuel economy increased from 19.5 km/L to 26 km/L for a 10 km/h increase in speed.
Now, we can continue this process to find the fuel economy rates for other 10 km/h intervals.
By comparing these rates, we can determine the speed that gives the best fuel economy, which will be the point where the fuel economy rate is highest within the given function.
To determine the rate of change in fuel economy for 10 km/h intervals in speed, we need to find the slope of the function f(x)=-0.005x^2+0.8x+12 at different values of x.
The slope of a function represents the rate of change. In this case, it represents the rate of change in fuel economy for a specific speed. To calculate the slope, we need to find the derivative of the function.
Taking the derivative of f(x)=-0.005x^2+0.8x+12 with respect to x, we get:
f'(x) = -0.01x + 0.8
This derivative equation gives us the rate of change in fuel economy (f'(x)) for a given speed (x).
Now, let's determine the rate of change in fuel economy for 10 km/h intervals in speed. We will calculate the slope at specific values of x, which correspond to 10 km/h intervals.
Let's start with the given speed intervals:
x1 = 0 km/h
x2 = 10 km/h
x3 = 20 km/h
and so on.
Calculate the corresponding fuel economy using the function f(x). For example, for x1 = 0 km/h, we have:
f(x1) = -0.005(0)^2+0.8(0)+12 = 12
Let's calculate the rates of change for the intervals:
f'(x1) = -0.01(0) + 0.8 = 0.8
f'(x2) = -0.01(10) + 0.8 = 0.7
f'(x3) = -0.01(20) + 0.8 = 0.6
Continue this calculation for each interval.
To determine the speed that gives the best fuel economy, we need to find the maximum value of the function f(x). This represents the peak fuel economy.
We can determine the maximum value by finding the x-coordinate of the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula -b/2a.
For the function f(x)=-0.005x^2+0.8x+12, a = -0.005 and b = 0.8.
x-coordinate of the vertex = -0.8 / (2 * -0.005) = 80
Therefore, the speed that gives the best fuel economy is 80 km/h.