How are the average rate of change and the instantaneous rate of change related for ƒ(x) = 2x + 5 ?
Since it is a linear function, that is, it can be represented by a straight line, they are the same.
The slope of the line is constant, so it is the same whether you take an average, or take the instantaneous rate of change.
To understand the relationship between the average rate of change and the instantaneous rate of change for the function ƒ(x) = 2x + 5, we need to first define each concept.
The average rate of change represents the overall change in the function output over a specific interval of the function's domain. It is determined by finding the slope of the secant line between two points on the function graph.
The formula to calculate the average rate of change is:
Average Rate of Change = (ƒ(x₂) - ƒ(x₁)) / (x₂ - x₁)
On the other hand, the instantaneous rate of change represents the rate of change at a specific point on the function graph. It is determined by finding the slope of the tangent line at that particular point.
To calculate the instantaneous rate of change, we need to take the derivative of the function. In this case, the derivative of ƒ(x) = 2x + 5 is simply the coefficient of x, which is 2.
Hence, the instantaneous rate of change for ƒ(x) = 2x + 5 is always 2.
Now, the relationship between the average rate of change and the instantaneous rate of change is that when the interval over which the average rate of change is calculated becomes infinitesimally small, it approaches the value of the instantaneous rate of change.
In our case, since the derivative is a constant 2, the instantaneous rate of change is always 2. As a result, regardless of the interval chosen, the average rate of change over that interval will always approach 2 as the interval becomes smaller.
Therefore, for the function ƒ(x) = 2x + 5, the average rate of change and the instantaneous rate of change are equal and constant, with both values being 2.