The Central Idea
When a quantity changes, we often want to know not only how much it changed, but how much it changed relative to something else.
If a quantity changes from an initial value to a final value, its change is
\[ \Delta y=y_2-y_1. \]If the corresponding input changes by
\[ \Delta x=x_2-x_1, \]then the average rate of change is
\[ \boxed{\frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}}. \]The numerator measures the change in the output. The denominator measures the change in the input. Their quotient tells us how much output change occurs, on average, for each unit of input change.
What Is a Rate?
A rate is a ratio that compares quantities with different roles, and often different units.
Examples include:
- miles per hour;
- dollars per item;
- liters per minute;
- people per square kilometer;
- degrees Celsius per kilometer of altitude;
- dollars of revenue per customer.
The word per usually signals division. A speed of 60 miles per hour means
\[ \frac{60\text{ miles}}{1\text{ hour}}. \]Change Is a Difference
Before computing a rate, we must compute a change. Change means final value minus initial value.
\[ \boxed{\text{change}=\text{final}-\text{initial}}. \]If a temperature rises from \(12^\circ\text{C}\) to \(19^\circ\text{C}\), then
\[ \Delta T=19-12=7^\circ\text{C}. \]If a bank balance falls from \(\$900\) to \(\$760\), then
\[ \Delta B=760-900=-140. \]The negative sign is meaningful: the balance decreased by \(\$140\).
The Delta Symbol
The uppercase Greek letter delta,
\[ \Delta, \]is commonly used to mean change in.
Thus,
\[ \Delta x=x_2-x_1 \]means the change in \(x\), while
\[ \Delta y=y_2-y_1 \]means the change in \(y\).
Delta notation will appear repeatedly in later work with slopes, finite differences, derivatives, and accumulated change.
Average Rate of Change
For a function \(y=f(x)\), the average rate of change from \(x=a\) to \(x=b\) is
\[ \boxed{ \frac{f(b)-f(a)}{b-a} }. \]This formula compares the total change in the function value with the total change in the input.
It does not require the function to change at the same rate throughout the interval.
Average Does Not Mean Arithmetic Mean Here
The phrase average rate of change can be misunderstood. We are not usually averaging a list of already-known rates.
Instead, we compute one quotient:
\[ \frac{\text{total output change}}{\text{total input change}}. \]For example, if a car travels 180 miles in 3 hours, its average speed is
\[ \frac{180\text{ miles}}{3\text{ hours}} = 60\text{ miles per hour}. \]The car does not need to have traveled at exactly 60 miles per hour during every moment of the trip.
Constant Rates
A rate is constant when equal changes in the input always produce proportional changes in the output.
Consider
\[ f(x)=3x+2. \]If \(x\) increases by 1, then \(f(x)\) increases by 3:
\[ f(x+1)-f(x) = [3(x+1)+2]-(3x+2) =3. \]The function has a constant rate of change of
\[ \boxed{3}. \]Linear Functions Have Constant Rate of Change
For a linear function
\[ f(x)=mx+c, \]the average rate of change over any interval is \(m\).
From \(x=a\) to \(x=b\), we have
\[ \frac{f(b)-f(a)}{b-a} = \frac{(mb+c)-(ma+c)}{b-a} = \frac{m(b-a)}{b-a} = \boxed{m}, \]provided \(a\ne b\).
A Constant Rate in a Table
Suppose a function is represented by the table
| \(x\) | \(f(x)\) |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 13 |
| 3 | 17 |
Each time \(x\) increases by 1, the output increases by 4.
\[ \frac{9-5}{1-0} = \frac{13-9}{2-1} = \frac{17-13}{3-2} =4. \]The repeated rate is evidence of a linear relationship.
Changing Rates
A rate is changing when the output does not change by the same amount for equal changes in the input.
Consider
\[ f(x)=x^2. \]Over successive one-unit intervals,
\[ f(1)-f(0)=1, \] \[ f(2)-f(1)=3, \] \[ f(3)-f(2)=5. \]The output changes by larger amounts as \(x\) increases. The rate is not constant.
A Changing Rate in a Table
| \(x\) | \(f(x)=x^2\) | Change in \(f(x)\) |
|---|---|---|
| 0 | 0 | — |
| 1 | 1 | 1 |
| 2 | 4 | 3 |
| 3 | 9 | 5 |
| 4 | 16 | 7 |
Because the input steps are all 1, the output differences
\[ 1,\qquad3,\qquad5,\qquad7 \]are also the average rates of change across those one-unit intervals.
Example: Average Rate of Change of a Linear Function
Find the average rate of change of
\[ f(x)=5x-7 \]from \(x=2\) to \(x=8\).
First evaluate the endpoints:
\[ f(2)=5(2)-7=3, \] \[ f(8)=5(8)-7=33. \]Then compute
\[ \frac{f(8)-f(2)}{8-2} = \frac{33-3}{6} = \boxed{5}. \]This matches the slope coefficient because the function is linear.
Example: Average Rate of Change of a Quadratic
Find the average rate of change of
\[ f(x)=x^2 \]from \(x=1\) to \(x=4\).
\[ f(1)=1, \] \[ f(4)=16. \]Therefore,
\[ \frac{f(4)-f(1)}{4-1} = \frac{16-1}{3} = \boxed{5}. \]The function does not have a constant rate of 5. Rather, 5 summarizes the change across the entire interval from 1 to 4.
The Same Function Can Have Different Average Rates
Continue with
\[ f(x)=x^2. \]From \(x=0\) to \(x=2\),
\[ \frac{4-0}{2-0}=2. \]From \(x=2\) to \(x=4\),
\[ \frac{16-4}{4-2}=6. \]From \(x=4\) to \(x=6\),
\[ \frac{36-16}{6-4}=10. \]The changing average rates reflect the curvature of the graph.
Rate of Change as Slope
The familiar slope formula is
\[ m=\frac{y_2-y_1}{x_2-x_1}. \]This is exactly the average-rate-of-change formula.
For two points on the graph of \(f\),
\[ (a,f(a)) \]and
\[ (b,f(b)), \]the slope between them is
\[ \boxed{ \frac{f(b)-f(a)}{b-a} }. \]The line through those two points is called a secant line. The next lesson develops this geometric viewpoint in greater detail.
Why Two Distinct Inputs Are Required
In
\[ \frac{f(b)-f(a)}{b-a}, \]the denominator cannot equal zero.
Therefore,
\[ \boxed{a\ne b}. \]If \(a=b\), then there is no input interval across which to measure average change, and the quotient would require division by zero.
Units of a Rate
The units of a rate come from dividing the output units by the input units.
If distance \(d\) is measured in miles and time \(t\) in hours, then
\[ \frac{\Delta d}{\Delta t} \]has units
\[ \boxed{\text{miles per hour}}. \]If cost is measured in dollars and production in units, then
\[ \frac{\Delta C}{\Delta q} \]has units of dollars per unit.
Carrying units through the calculation is one of the best ways to check whether a rate has been interpreted correctly.
Example: Average Speed
A vehicle is 45 miles from its starting point after 1 hour and 225 miles from its starting point after 4 hours.
The change in distance is
\[ 225-45=180\text{ miles}. \]The change in time is
\[ 4-1=3\text{ hours}. \]The average rate of change of distance with respect to time is
\[ \boxed{ \frac{180\text{ miles}}{3\text{ hours}} =60\text{ miles per hour} }. \]Average Velocity and Direction
When position can increase or decrease along a directed line, the sign of the rate matters.
Suppose position changes from 30 meters to 6 meters in 4 seconds.
\[ \Delta s=6-30=-24\text{ m}. \]Therefore,
\[ \frac{\Delta s}{\Delta t} = \frac{-24}{4} = \boxed{-6\text{ m/s}}. \]The negative sign indicates motion in the negative direction. It does not mean that the object has a negative amount of speed.
Positive, Negative, and Zero Rates
The sign of an average rate communicates direction of change.
- A positive rate means the output increased as the input increased.
- A negative rate means the output decreased as the input increased.
- A zero rate means the output had the same value at both endpoints.
A zero average rate does not necessarily mean the function remained constant throughout the interval. It only means the endpoint values were equal.
Zero Average Rate Does Not Imply No Change
Consider
\[ f(x)=x^2 \]on the interval from \(x=-2\) to \(x=2\).
\[ f(-2)=4 \]and
\[ f(2)=4. \]Thus,
\[ \frac{f(2)-f(-2)}{2-(-2)} = \frac{4-4}{4} = \boxed{0}. \]Yet the function decreases from 4 to 0 and then increases back to 4. The average rate summarizes the net endpoint change, not every event inside the interval.
Average Rate from a Table
Suppose a population is recorded as follows:
| Year | Population |
|---|---|
| 2020 | 12,400 |
| 2022 | 13,100 |
| 2025 | 14,450 |
From 2020 to 2025, the average annual rate of change is
\[ \frac{14{,}450-12{,}400}{2025-2020} = \frac{2050}{5} = \boxed{410\text{ people per year}}. \]We use the endpoint values required by the interval. Intermediate data are not needed for this particular average-rate calculation.
Unequal Input Intervals
When input intervals have different widths, raw output differences cannot be compared fairly by themselves.
Suppose sales increase by \(\$600\) over 2 months and later increase by \(\$900\) over 6 months.
The first average rate is
\[ \frac{600}{2}=300\text{ dollars per month}. \]The second is
\[ \frac{900}{6}=150\text{ dollars per month}. \]Even though the second total increase is larger, its average rate is smaller.
Average Rate from a Formula
Suppose
\[ C(q)=50+8q+q^2 \]gives a cost, in dollars, associated with producing \(q\) units.
Find the average rate of change from \(q=10\) to \(q=14\).
\[ C(10)=50+80+100=230, \] \[ C(14)=50+112+196=358. \]Therefore,
\[ \frac{C(14)-C(10)}{14-10} = \frac{358-230}{4} = \boxed{32\text{ dollars per unit}}. \]Function Notation Keeps the Structure Visible
The average-rate formula
\[ \frac{f(b)-f(a)}{b-a} \]is useful because it separates the calculation into four pieces:
- the later or second input, \(b\);
- the earlier or first input, \(a\);
- the corresponding output values, \(f(b)\) and \(f(a)\);
- the quotient of the two changes.
Keeping this structure visible reduces sign errors and helps prepare for later difference-quotient notation.
Using an Increment \(h\)
Another useful way to describe an interval is to begin at \(x\) and move by an increment \(h\).
The second input is then
\[ x+h. \]The average rate of change across that interval is
\[ \frac{f(x+h)-f(x)}{(x+h)-x} = \boxed{\frac{f(x+h)-f(x)}{h}}, \]where \(h\ne0\).
This expression is called a difference quotient and will be studied directly later in this section.
Example: Building a Difference Quotient
Let
\[ f(x)=x^2. \]Then
\[ f(x+h)=(x+h)^2. \]The average rate from \(x\) to \(x+h\) is
\[ \frac{(x+h)^2-x^2}{h}. \]Expand the numerator:
\[ \frac{x^2+2xh+h^2-x^2}{h} = \frac{2xh+h^2}{h}. \]For \(h\ne0\),
\[ \boxed{2x+h}. \]This tells us that the average rate of \(x^2\) across an interval of width \(h\) beginning at \(x\) depends on both the starting point and the interval width.
Finite Differences
A finite difference is the difference between function values at distinct inputs.
For a step size \(h\), a first difference can be written
\[ \Delta f=f(x+h)-f(x). \]Dividing by the step size gives the corresponding average rate:
\[ \boxed{ \frac{\Delta f}{\Delta x} = \frac{f(x+h)-f(x)}{h} }. \]Finite differences provide an important connection between tables, algebra, numerical methods, and the later study of derivatives.
Equal-Step Data
When inputs are equally spaced, differences can be compared directly.
| \(x\) | \(f(x)\) | First difference |
|---|---|---|
| 0 | 2 | — |
| 2 | 8 | 6 |
| 4 | 14 | 6 |
| 6 | 20 | 6 |
The output increases by 6 whenever the input increases by 2, so the rate is
\[ \frac{6}{2}=\boxed{3}. \]Constant first differences with equal input spacing indicate a constant rate of change.
Do Not Confuse Difference with Rate
A difference measures a change in one quantity. A rate compares that change with a change in another quantity.
If output rises by 12 while input rises by 3, then
\[ \Delta y=12, \]but the rate is
\[ \frac{\Delta y}{\Delta x} = \frac{12}{3} = \boxed{4}. \]The value 12 is the output change. The value 4 is the output change per unit of input change.
Rate Depends on Which Quantity Is in the Denominator
Suppose a machine produces 240 parts in 6 hours.
Parts per hour is
\[ \frac{240\text{ parts}}{6\text{ hours}} = 40\text{ parts per hour}. \]Hours per part is the reciprocal comparison:
\[ \frac{6\text{ hours}}{240\text{ parts}} = 0.025\text{ hour per part}. \]Both ratios are valid, but they answer different questions.
Interpreting a Rate in Context
A numerical rate should be stated with its quantities and units.
Suppose the average rate of change of revenue with respect to customers is
\[ 42. \]If revenue is measured in dollars and customers are counted as people, a useful interpretation is:
Over the stated interval, revenue increased by an average of approximately \(\$42\) for each additional customer.
The phrase over the stated interval matters when the underlying rate is not constant.
Rate as Sensitivity
Rate of change can also be understood as a measure of sensitivity.
If a small change in \(x\) is associated with a large change in \(f(x)\), then the output is highly sensitive to the input over that interval.
If the output changes very little while the input changes substantially, the relationship has a small average rate over that interval.
This viewpoint becomes useful in modeling, optimization, economics, engineering, and numerical analysis.
Average Rate and Curve Shape
For a straight line, every secant slope is the same. For a curve, secant slopes can change as the chosen interval changes.
This gives us an early way to investigate a graph:
- positive average rates suggest net increase;
- negative average rates suggest net decrease;
- larger positive rates suggest steeper net increase;
- more negative rates suggest steeper net decrease;
- changing rates signal nonlinearity or changing behavior.
Later lessons refine these ideas from interval-based averages to local behavior.
Average Rate Is Interval-Dependent
For nonlinear functions, an average rate should always be associated with a stated interval.
Saying only
“the average rate is 7”
is incomplete unless we know the interval and the relevant quantities.
A more complete statement is:
“From \(x=2\) to \(x=5\), the average rate of change of \(f\) is 7 units of output per unit of input.”
Example: A Decreasing Function
Let
\[ f(x)=20-2x^2. \]Find the average rate of change from \(x=1\) to \(x=3\).
\[ f(1)=20-2=18, \] \[ f(3)=20-18=2. \]Therefore,
\[ \frac{2-18}{3-1} = \frac{-16}{2} = \boxed{-8}. \]On average, the function decreases by 8 output units for each one-unit increase in \(x\) across this interval.
Example: A Rate from Real Data
Suppose the water level in a reservoir is 82.4 meters at 8:00 a.m. and 80.9 meters at 2:00 p.m.
The change in water level is
\[ 80.9-82.4=-1.5\text{ m}. \]The elapsed time is
\[ 6\text{ h}. \]The average rate is
\[ \frac{-1.5}{6} = \boxed{-0.25\text{ m/h}}. \]The negative sign tells us that the water level fell over the interval.
Rates from Graphs
A graph can be used to estimate an average rate of change by reading two points and computing their slope.
If a graph appears to contain the points
\[ (2,5) \]and
\[ (7,17), \]then the estimated average rate is
\[ \frac{17-5}{7-2} = \frac{12}{5} = \boxed{2.4}. \]If the coordinates are estimated from a drawing, the resulting rate is also an estimate.
Exact Data and Approximate Data
Rates may be exact or approximate depending on the source values.
If a formula gives exact endpoint values, the resulting rate can often be expressed exactly.
If measurements are rounded, or graph coordinates are estimated visually, the rate should usually be reported with suitable precision.
The precision of a rate should not imply more certainty than the underlying data support.
Rate Versus Percentage Change
An ordinary rate of change and a percentage change answer different questions.
If a quantity rises from 100 to 120 over 4 years, the average absolute rate is
\[ \frac{120-100}{4} = 5\text{ units per year}. \]The total percentage change is
\[ \frac{120-100}{100}\cdot100\% =20\%. \]One compares change with elapsed input; the other compares change with the initial amount.
Constant Rate Versus Constant Percentage Rate
A quantity can grow at a constant absolute rate or at a constant percentage rate, but these are different patterns.
Adding \(\$100\) each year is constant absolute change.
Growing by 5% each year is constant relative or percentage growth.
Constant absolute change is associated with linear behavior. Constant percentage growth is associated with exponential behavior.
Distinguishing these patterns is important when interpreting financial, biological, and population models.
Average Rate Does Not Reveal Every Local Detail
Two functions can have the same endpoint values and therefore the same average rate across an interval while behaving very differently inside it.
Average rate is therefore a useful summary, but not a complete description of local behavior.
This limitation motivates the next question:
How can we describe the rate of change near one particular point rather than across a broad interval?
That question leads to secant lines, tangent lines, and local rate.
Common Error: Reversing Only One Difference
Consider the points
\[ (2,5) \]and
\[ (6,17). \]A correct calculation is
\[ \frac{17-5}{6-2} =3. \]Reversing both differences also works:
\[ \frac{5-17}{2-6} =3. \]But reversing only one difference changes the sign incorrectly.
Common Error: Forgetting the Input Change
Suppose an output rises from 40 to 64 while the input rises from 3 to 7.
The output change is
\[ 64-40=24, \]but the rate is not 24.
The input change is
\[ 7-3=4, \]so the average rate is
\[ \boxed{\frac{24}{4}=6}. \]Common Error: Averaging Endpoint Outputs
For \(f(2)=5\) and \(f(6)=17\), the arithmetic mean of the endpoint outputs is
\[ \frac{5+17}{2}=11. \]But 11 is not the average rate of change.
The average rate is
\[ \frac{17-5}{6-2} = \boxed{3}. \]Common Error: Ignoring Units
A result such as
\[ 12 \]may be mathematically incomplete in context.
A better result might be
\[ 12\text{ dollars per hour} \]or
\[ 12\text{ kilometers per liter}. \]Units identify what the quotient actually measures.
Common Error: Assuming an Average Rate Is Constant Everywhere
If \(f(x)=x^2\) has an average rate of 5 from \(x=1\) to \(x=4\), this does not mean that its rate is 5 throughout the interval.
The function is nonlinear, and its rate changes with \(x\).
The value 5 is an interval summary.
Common Error: Treating Equal Endpoint Values as a Constant Function
If
\[ f(a)=f(b), \]then the average rate from \(a\) to \(b\) is zero.
That does not prove that
\[ f(x) \]remained constant between the endpoints.
A curve may rise and fall while ending at the same output value.
A Reliable Procedure for Average Rate of Change
- Identify the input variable and output quantity.
- Identify the two input values that define the interval.
- Find the corresponding output values.
- Compute output change as final output minus initial output.
- Compute input change as final input minus initial input.
- Divide output change by input change.
- Simplify the quotient.
- Attach the correct units.
- Interpret the sign and magnitude in context.
- State the interval when the rate is not globally constant.
Example: Full Workflow from a Formula
Let
\[ P(t)=2t^2+10t+50, \]where \(P\) is measured in units of production and \(t\) in hours. Find the average rate from \(t=2\) to \(t=5\).
Evaluate the endpoints:
\[ P(2)=2(4)+20+50=78, \] \[ P(5)=2(25)+50+50=150. \]Output change:
\[ \Delta P=150-78=72. \]Input change:
\[ \Delta t=5-2=3. \]Average rate:
\[ \boxed{ \frac{\Delta P}{\Delta t} = \frac{72}{3} =24\text{ production units per hour} }. \]Example: Comparing Two Intervals
Continue with
\[ P(t)=2t^2+10t+50. \]From \(t=0\) to \(t=2\),
\[ \frac{P(2)-P(0)}{2-0} = \frac{78-50}{2} =14. \]From \(t=2\) to \(t=5\), we found
\[ 24. \]The later interval has the larger average rate. This is consistent with a process whose rate is increasing as time increases.
Average Rates Can Be Used for Estimation
If a quantity changes approximately linearly over a short interval, an average rate can provide a practical estimate.
Suppose a measured quantity is increasing at an average rate of about
\[ 3.2\text{ units per hour}. \]Over an additional 2 hours, a rough linear estimate of the change is
\[ 3.2(2)=6.4\text{ units}. \]Such an estimate assumes that the recent average rate remains reasonably representative over the new interval.
From Average Rate to Local Rate
Average rate measures change across an interval with nonzero width.
To investigate behavior near a single input, we can compare rates over progressively shorter intervals.
For example, near \(x=2\) we might examine intervals such as
\[ [2,3],\qquad[2,2.5],\qquad[2,2.1]. \]The corresponding secant slopes can reveal how the function is behaving near \(x=2\).
This is the conceptual bridge to tangent lines and local rate of change.
Practice 1: Constant Rate from a Table
Determine whether the rate is constant and, if so, find it.
| \(x\) | \(f(x)\) |
|---|---|
| 0 | 7 |
| 2 | 13 |
| 4 | 19 |
| 6 | 25 |
Practice 2: Average Rate of a Linear Function
Find the average rate of change of
\[ f(x)=7x+4 \]from \(x=-1\) to \(x=5\).
Practice 3: Average Rate of a Quadratic
Find the average rate of change of
\[ f(x)=x^2+2x \]from \(x=1\) to \(x=4\).
Practice 4: Negative Rate
A tank contains 920 liters of water at noon and 740 liters at 3:00 p.m. Find the average rate of change of water volume with respect to time.
Practice 5: Unequal Time Interval
A business has 1,240 subscribers in March and 1,690 subscribers in September. Find the average monthly change over the six-month interval.
Practice 6: Zero Average Rate
Find the average rate of change of
\[ f(x)=x^2-4x \]from \(x=1\) to \(x=3\). Explain what the result does and does not tell you.
Practice 7: Difference Quotient
For
\[ f(x)=3x^2, \]simplify
\[ \frac{f(x+h)-f(x)}{h} \]for \(h\ne0\).
Practice 8: Rate from Two Data Points
A temperature is \(18^\circ\text{C}\) at an altitude of 500 meters and \(11^\circ\text{C}\) at an altitude of 1,500 meters. Find the average rate of temperature change with respect to altitude.
Practice 9: Compare Intervals
For
\[ f(x)=x^2, \]find the average rate of change on \([1,2]\) and on \([4,5]\). Which interval has the greater average rate?
Practice 10: Identify the Error
A student computes the average rate of change from \((2,5)\) to \((6,17)\) as
\[ \frac{17-5}{2-6}=-3. \]Explain the error and give the correct rate.
Practice Answers
Practice 1
Each output increase is 6 while each input increase is 2:
\[ \frac{6}{2}=3. \]Therefore, the rate is constant:
\[ \boxed{3}. \]Practice 2
\[ f(-1)=-3, \] \[ f(5)=39. \] \[ \frac{39-(-3)}{5-(-1)} = \frac{42}{6} = \boxed{7}. \]Practice 3
\[ f(1)=3, \] \[ f(4)=24. \] \[ \frac{24-3}{4-1} = \frac{21}{3} = \boxed{7}. \]Practice 4
The water-volume change is
\[ 740-920=-180\text{ L}. \]Over 3 hours,
\[ \boxed{-60\text{ L/h}}. \]Practice 5
\[ \frac{1690-1240}{6} = \frac{450}{6} = \boxed{75\text{ subscribers per month}}. \]Practice 6
\[ f(1)=1-4=-3, \] \[ f(3)=9-12=-3. \] \[ \frac{-3-(-3)}{3-1}=0. \]Therefore,
\[ \boxed{0}. \]The endpoint values are equal, so the net average rate is zero. This does not mean the function remained constant between \(x=1\) and \(x=3\).
Practice 7
\[ f(x+h)=3(x+h)^2 =3x^2+6xh+3h^2. \] \[ \frac{f(x+h)-f(x)}{h} = \frac{6xh+3h^2}{h} = \boxed{6x+3h}. \]Practice 8
\[ \frac{11-18}{1500-500} = \frac{-7}{1000} = \boxed{-0.007^\circ\text{C per meter}}. \]Equivalently, the average change is
\[ \boxed{-7^\circ\text{C per 1000 meters}}. \]Practice 9
On \([1,2]\),
\[ \frac{4-1}{2-1}=3. \]On \([4,5]\),
\[ \frac{25-16}{5-4}=9. \]Therefore, the second interval has the greater average rate:
\[ \boxed{9>3}. \]Practice 10
The numerator uses final minus initial, but the denominator uses initial minus final. The subtraction order must be consistent.
\[ \frac{17-5}{6-2} = \frac{12}{4} = \boxed{3}. \]What to Remember
A rate compares change in one quantity with change in another:
\[ \boxed{ \text{average rate of change} = \frac{\Delta y}{\Delta x} }. \]For a function \(f\), from \(x=a\) to \(x=b\),
\[ \boxed{ \frac{f(b)-f(a)}{b-a} }, \qquad a\ne b. \]- linear functions have constant rate of change;
- nonlinear functions generally have rates that change from interval to interval;
- the sign of a rate communicates direction of net change;
- the units are output units per input unit;
- average rate summarizes an interval and does not describe every local detail;
- the average rate is the slope of the secant line through the interval endpoints.
Where We Go Next
Average rate of change measures the slope between two distinct points. To analyze a curve more closely, we next ask what happens as those two points are brought nearer together.
The next lesson, Secant Lines, Tangent Lines, and Local Rate, develops the geometric connection between average change across an interval and the local behavior of a curve near a single point.