Why the Chosen Interval Matters
For a nonlinear function, an average rate describes change across a particular interval. Change the interval, and the average rate may change too. The rate is not a single label attached to the whole function: it summarizes what happened between two selected inputs.
Nested intervals make this dependence especially clear. We can keep one endpoint fixed and move the other, or compare a smaller interval lying inside a larger one. A table shows how the computed values change; a graph shows which portions of the function each value summarizes.
This is useful whenever a reported rate depends on a time period, input range, or measurement window. A rate averaged over a long span may combine quite different behavior within that span. A shorter span reports change over a more restricted part of the function.
Average Rates on Nested Intervals
An interval \([u,v]\) is nested inside \([a,b]\) when \(a\leq u<v\leq b\). Both intervals have positive width, and the smaller one lies entirely within the larger one. In a common and convenient arrangement, the intervals share a starting point: \([a,u]\) lies inside \([a,b]\), where \(a<u<b\).
For a linear function, average rates are the same on every interval in its domain. A nonlinear function does not generally have this property. Different intervals can include different parts of a curve, so their secant slopes can differ. In particular, a larger interval can combine parts where the function is increasing, decreasing, or changing at different rates.
Worked Example: A Quadratic Function on Shared-Endpoint Intervals
Worked Example: A Quadratic Function on Shared-Endpoint Intervals
Let \(f(x)=x^2\). Find the average rates on the nested intervals \([0,2]\), \([0,1]\), and \([0,\frac12]\).
First evaluate the function at the endpoints needed:
On \([0,2]\), the output change is \(4-0=4\), and the input change is \(2-0=2\). Thus the average rate is
On \([0,1]\), the output change is \(1-0=1\), and the input change is \(1-0=1\):
On \([0,\frac12]\), the output change is \(\frac14-0=\frac14\), and the input change is \(\frac12-0=\frac12\):
The calculations can be organized as follows:
| Interval | Output change | Input change | Average rate |
|---|---|---|---|
| \([0,2]\) | \(4-0=4\) | \(2-0=2\) | \(4/2=2\) |
| \([0,1]\) | \(1-0=1\) | \(1-0=1\) | \(1/1=1\) |
| \([0,\frac12]\) | \(\frac14-0=\frac14\) | \(\frac12-0=\frac12\) | \((\frac14)/(\frac12)=\frac12\) |
Each rate checks against the definition: multiplying the rate by the interval width gives the output change. For example, \(2(2)=4\), \(1(1)=1\), and \(\frac12(\frac12)=\frac14\), matching the respective changes in \(f\).
On a graph of \(y=x^2\), each rate is the slope of a secant from \((0,0)\) to the point at the other endpoint. The three secants use different portions of the curve. Their different slopes show that the average rate reported for this nonlinear function depends on the selected interval.
Reading Nested-Interval Rates from a Table
A table can supply the function values directly, without giving a formula. For each interval, select its two endpoint values, subtract in the same order in the numerator, and divide by the interval width. A table of rates is useful because it keeps the interval associated with each reported value explicit.
Worked Example: A Table with Increasing and Decreasing Behavior
Worked Example: A Table with Increasing and Decreasing Behavior
Suppose a function is described by the values below. Compute its average rates on \([0,3]\), \([0,2]\), and \([0,1]\), which are nested intervals with a shared starting point.
| \(x\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| \(f(x)\) | 0 | \(-2\) | 2 | 18 |
For \([0,3]\), \(f(3)-f(0)=18-0=18\), and \(3-0=3\). The average rate is \(18/3=6\).
For \([0,2]\), \(f(2)-f(0)=2-0=2\), and \(2-0=2\). The average rate is \(2/2=1\).
For \([0,1]\), \(f(1)-f(0)=-2-0=-2\), and \(1-0=1\). The average rate is \(-2/1=-2\).
| Interval | Calculation | Average rate |
|---|---|---|
| \([0,3]\) | \((18-0)/(3-0)=18/3\) | 6 |
| \([0,2]\) | \((2-0)/(2-0)=2/2\) | 1 |
| \([0,1]\) | \((-2-0)/(1-0)=-2/1\) | \(-2\) |
Each answer reproduces the corresponding output change when multiplied by the interval width: \(6(3)=18\), \(1(2)=2\), and \((-2)(1)=-2\). The negative rate on \([0,1]\) records that the output decreases between those endpoints. The positive rates on the larger intervals summarize the net change between their own endpoints.
If the graph connects these tabulated points, it falls from \((0,0)\) to \((1,-2)\) and then rises through \((2,2)\) and \((3,18)\). The secant over \([0,3]\) summarizes the whole span, including both the fall and rise. The secant over \([0,1]\) records only the initial fall. Thus, one outer-interval rate can conceal variation that is visible in the table and graph.
Why an Outer Rate Combines Smaller-Interval Rates
There is a precise reason an average rate over an interval reflects the rates on its parts. Consider three inputs \(a<u<b\). The interval \([a,b]\) is divided into \([a,u]\) and \([u,b]\). Let \(r_1\) be the average rate on the first part and \(r_2\) the average rate on the second:
Rearranging these definitions gives the output changes on the two parts. Adding those changes recovers the change across the entire interval:
Now divide by the full width \(b-a=(u-a)+(b-u)\). The result is:
This derivation explains why the wider interval does not simply report one of its smaller-interval rates. It combines the output changes from both pieces, with each piece contributing according to its width. If the rates on the pieces differ, the outer interval summarizes both of them.
Worked Example: A Nested Interval Inside a Larger One
Worked Example: A Nested Interval Inside a Larger One
Let \(f(x)=x^3\). Compare the average rate on the inner interval \([1,3]\) with the rate on the outer interval \([0,4]\), then verify the weighted-rate result using the pieces \([0,1]\), \([1,3]\), and \([3,4]\).
First calculate the needed function values:
The average rate on the outer interval is
The average rate on the inner interval is
For the two remaining pieces, the rates are \((1-0)/(1-0)=1\) on \([0,1]\) and \((64-27)/(4-3)=37\) on \([3,4]\). The outer rate formed from all three pieces is therefore
This agrees with the direct calculation on \([0,4]\). As an endpoint check, the weighted output changes add to \(1+26+37=64\), which equals \(f(4)-f(0)\). The example also shows that an inner interval can have a rate different from the outer interval, even when it is fully contained within it.
A Common Mistake: Dividing by the Wrong Width
When intervals are nested, it is easy to use the output change from one interval and the width from another. For \(f(x)=x^2\), the rate on \([0,1]\) uses \(f(1)-f(0)=1-0=1\). A mistaken calculation might divide this change by the outer interval's width, \(2\), and report
This is incorrect for \([0,1]\): its width is \(1-0=1\), not \(2\). The corrected calculation is
The incorrect value \(\frac12\) mixes the numerator from \([0,1]\) with the denominator from \([0,2]\). To check any average rate, multiply it by the width of the interval named in the calculation. For the correct rate, \(1(1)=1\), the actual output change on \([0,1]\).
Using Tables and Graphs Carefully
A table gives endpoint values and supports exact calculations when those values are exact. A graph gives a visual estimate of the secant slope when precise values are unavailable. In either case, the rate belongs to its specified interval. Record the endpoints along with the value so that a rate is not mistaken for a general property of the function.
- For each interval, identify both endpoint inputs before calculating.
- Use the output change in the same endpoint order as the input change.
- Label each rate with its interval; nested intervals may produce distinct values.
- Use the curve between endpoints to interpret what behavior an outer rate combines.
A graph can suggest where a function changes its behavior, but a secant slope uses only the two chosen endpoints. It does not describe every rise and fall between them. The table example illustrates this: the rate on \([0,3]\) is positive even though the graph initially falls, because the endpoint output at \(3\) is much greater than at \(0\).
Why Interval Choice Matters
Average rates are often reported as summaries: average output change per unit of input over a stated range. The range is part of the meaning of the number. If two reports use different intervals, their values may differ because each includes a different portion of the function's behavior—not because either calculation is necessarily wrong.
For nonlinear functions, comparing nested intervals helps reveal this dependence. A table can show how endpoint-based rates vary as the interval changes, while a graph can show which parts of the curve contribute to each summary. These are descriptions of average change over finite intervals; they do not by themselves assert a rate at a single input.
Check Your Understanding
Use the definition of average rate and keep the interval attached to each calculation.
- For \(f(x)=x^2\), calculate and compare the average rates on \([1,3]\) and its nested interval \([1,2]\).
- A table gives \(f(0)=5\), \(f(2)=9\), and \(f(4)=13\). Find the average rates on \([0,4]\) and \([0,2]\).
- In your own words, explain why an outer interval can combine different behavior from smaller intervals inside it.
- State the average-rate formula and identify the quantity that must appear in its denominator.
- A calculation uses an output change from an interval of width \(3\) but divides by \(5\). Explain why the interval and denominator must be checked before accepting the result.
