Unit 1 · Topic 1.2

Rates of Change

Measure how much output changes per unit of input, interpret that number in context, and estimate how rapidly a function is changing near a specific point.

Learning Goals

Distinguish an output change from an average rate of change.
Calculate average rate of change from a table, formula, graph, or verbal description.
Connect average rate of change to the slope of a secant line.
Use averages over small intervals to approximate and compare rates at two points.
Interpret the sign, magnitude, and units of a rate in context.

1. Change Versus Rate of Change

Over the input interval from \(a\) to \(b\), the output change is \(f(b)-f(a)\). A rate compares that output change with the corresponding input change.

\[\Delta y=f(b)-f(a),\qquad \Delta x=b-a\]

If distance increases by 120 miles, that is a change. If it increases by 120 miles over 3 hours, the average rate is 40 miles per hour.

2. Average Rate of Change

The average rate of change on \([a,b]\) is the constant rate that would produce the same total output change over the same input interval.

\[\operatorname{AROC}_{[a,b]}=\frac{f(b)-f(a)}{b-a},\qquad a\ne b\]

Read it as “change in output per one unit of input.” Its units are always output units per input unit.

3. Calculate from a Table

The table records the amount \(W(t)\), in liters, after \(t\) minutes.

Time \(t\) (min)	0	2	5	8
Amount \(W(t)\) (L)	10	18	24	15

Interval	Calculation	Average rate	Interpretation
\([0,2]\)	\((18-10)/(2-0)\)	4 L/min	The amount increases by an average of 4 liters each minute.
\([2,5]\)	\((24-18)/(5-2)\)	2 L/min	The amount is still increasing, but more slowly on this interval.
\([5,8]\)	\((15-24)/(8-5)\)	\(-3\) L/min	The amount decreases by an average of 3 liters each minute.

4. Graphical Meaning: Secant Slope

The average rate of change is the slope of the secant line through the interval endpoints; smaller nearby intervals can estimate the graph's steepness at a point.

The points \((a,f(a))\) and \((b,f(b))\) determine a secant line. Its slope equals the function's average rate of change on \([a,b]\).

\[m_{\text{secant}}=\frac{f(b)-f(a)}{b-a}\]

A steeper positive secant has a larger positive rate. A downward secant has a negative rate. A horizontal secant has average rate zero.

5. Worked Formula Example

For \(f(x)=x^2+2x\), find the average rate of change on \([1,4]\).

Evaluate the endpoints: \(f(1)=3\) and \(f(4)=24\).
Find the changes: \(\Delta y=24-3=21\) and \(\Delta x=4-1=3\).
Divide in the same order: \(21/3=7\).

The output increases by an average of 7 units for each one-unit increase in \(x\) over this interval.

6. Approximate the Rate at a Point

A single point does not provide two values for a slope calculation. Instead, use a small interval containing the point. As the interval narrows, its average rate can approximate the function's rate at that point.

For \(p(x)=x^2\), use intervals that begin at \(x=3\) and become progressively narrower:

Interval around 3	Average-rate calculation	Approximation
\([3,4]\)	\((16-9)/(4-3)\)	7
\([3,3.5]\)	\((12.25-9)/(3.5-3)\)	6.5
\([3,3.1]\)	\((9.61-9)/(3.1-3)\)	6.1

The averages approach 6 as the interval narrows, indicating that the rate at \(x=3\) is approximately 6. Topic 1.2 uses numerical approximation; a formal derivative is not required.

7. Compare Rates at Two Points

Use small intervals of the same width when possible. For \(p(x)=x^2\):

\[\operatorname{AROC}_{[0.9,1.1]}=2,\qquad \operatorname{AROC}_{[2.9,3.1]}=6\]

The function is changing more rapidly near \(x=3\) than near \(x=1\). On the graph, this appears as greater steepness near \(x=3\).

8. Interpret Sign, Magnitude, and Units

Positive rate: the two quantities change in the same direction.
Negative rate: as the input increases, the output decreases.
Zero average rate: the endpoints have the same output, although the function may change between them.
Larger absolute value: more output change per input unit.

A rate of \(-3\) °C/hour means temperature decreases by an average of 3 °C for each hour, not that the temperature itself equals \(-3\) °C.

9. Context Example

The temperature of a sample is modeled by \(T(t)=20+6t-t^2\), where \(t\) is hours and \(T\) is °C.

\[\operatorname{AROC}_{[0,2]}=\frac{28-20}{2}=4\text{ °C/hour}\]

\[\operatorname{AROC}_{[2,5]}=\frac{25-28}{3}=-1\text{ °C/hour}\]

The sample warms on average over the first interval and cools on average over the second. The interval must be named because one model can have different rates on different intervals.

10. AP Reasoning and Common Errors

A complete response should state the interval, rate, units, direction, and contextual meaning.

Response pattern: “On \([a,b]\), the average rate of change is ___ output units per input unit, so the output increases/decreases by an average of ___ for each input unit.”

Do not report only \(f(b)-f(a)\); that is output change, not rate.
Do not reverse the denominator while keeping the numerator order unchanged.
Do not omit units or confuse output units with rate units.
Do not claim the function changed at a constant rate throughout a nonlinear interval; the average is an equivalent constant rate.
Do not use a wide interval to describe a highly local rate when closer values are available.

Key Takeaways

Calculate and interpret average rates of change over intervals, then use small-interval averages to approximate and compare rates of change at points.
Core relationship: \(\operatorname{AROC}_{[a,b]}=\frac{f(b)-f(a)}{b-a}\)
Error check: Keep input order consistent, include units, and interpret the sign; an output difference alone is not a rate.

Checkpoint · Topic 1.2

Let \(R(t)=t^2+4t+10\), where \(t\) is measured in minutes and \(R\) is measured in milligrams.

Find and interpret the average rate of change on \([1,4]\), including units.
Approximate the rates at \(t=1\) and \(t=4\) using the intervals \([0.9,1.1]\) and \([3.9,4.1]\).
Determine where the graph is steeper and justify the comparison numerically.
Explain why the output change \(R(4)-R(1)\) is not itself an average rate.