Average Rate of Change

For any function f defined on an interval [a, b], the average rate of change equals (f(b) − f(a)) / (b − a). This single fraction is the slope of the secant line through the two points (a, f(a)) and (b, f(b)). You are not solving for a variable; you are dividing the net change in the output by the length of the interval. The result always carries units: if f is distance in metres and x is time in seconds, the average rate of change comes out in metres per second. If f is profit in dollars and x is years, the units are dollars per year.

The formula is identical to the difference quotient when you set b = a + h. Replace b with a + h and the expression becomes (f(a + h) − f(a)) / h. That is the starting point for the limit definition of the derivative. The only difference is that the average rate of change keeps h as a fixed number, while the derivative lets h approach zero to find an instantaneous rate of change. OpenStax Calculus Volume 1, section 3.1 "Defining the Derivative" uses this exact progression. Stewart Calculus sections 1.4 and 2.7-2.8 follow the same route from secant lines to tangent lines.

Most errors happen in the algebra of the numerator. You must treat (x + h) as a single input: f(x + h) = 1/(x + h), not 1/x + h, and f(x + h) = √(x + h), not √x + h. The increment h is not a number you pick; it cancels algebraically from the numerator before any limit is taken. If it does not cancel, the function is not differentiable at that x.

The average rate of change formula (f(b) − f(a)) / (b − a) works for any function for which f(a) and f(b) are defined. It is the slope of the secant line across the interval. OpenStax Precalculus 1.3 "Rates of Change" treats this as the core definition: the change in the output divided by the change in the input.

What the Numerator and Denominator Mean

f(b) − f(a) is the net vertical change. b − a is the horizontal span. The division gives the average rate at which the output changes per unit of input. If the result is positive, the function increased on average; if negative, it decreased; if zero, the function started and ended at the same value.

Link to the Difference Quotient

Let b = a + h. Then the average rate of change becomes (f(a + h) − f(a)) / h. This is the difference quotient. The increment h is the length of the interval. Stewart Calculus sections 2.7-2.8 uses this form to connect average rates of change to the derivative: take the limit as h → 0 and you get the instantaneous rate of change.

When you have a table of values, you do not need the function rule. Pick the two x-values that define your interval, read the corresponding f(x) values, and apply the formula. The algebra is identical: (f(b) − f(a)) / (b − a).

Worked Example: Population Table

The table shows the population P(t) of a town in thousands, where t is years since 2010.

t	P(t)
0	12
2	15
5	21

Find the average rate of change from t = 0 to t = 5. Use (P(5) − P(0)) / (5 − 0) = (21 − 12) / 5 = 9 / 5 = 1.8. The units are thousands of people per year. The population grew on average by 1,800 people per year over those five years. The same process works for any table with two matching entries.

From a graph, read the coordinates of two points on the curve that correspond to the endpoints of your interval. Then compute the slope of the secant line that connects them. That slope is the average rate of change. You do not need the equation of the curve.

Worked Example: Parabola Graph

On the graph of f(x) = x², the points (1, 1) and (4, 16) are on the curve. The average rate of change on [1, 4] is (16 − 1) / (4 − 1) = 15 / 3 = 5. The secant line through those points has slope 5. Visually, the line is steeper than the tangent at x = 1 (which has slope 2) but less steep than the tangent at x = 4 (which has slope 8). The average rate of change sits between the two instantaneous rates.

When the Graph Is Not a Function

The graph must pass the vertical line test for the average rate of change to be defined. A circle or a vertical line has multiple y-values for a single x, so the formula does not apply directly. In those cases you may need to treat each branch separately.

When you have the function rule, you evaluate f(a) and f(b) by substitution, then simplify the fraction. This is the most common form in calculus textbook problems.

Worked Example: Rational Function

Let f(x) = 1/x. Find the average rate of change on [2, 5].

Step 1: f(2) = 1/2, f(5) = 1/5.

Step 2: (1/5 − 1/2) / (5 − 2) = (2/10 − 5/10) / 3 = (−3/10) / 3 = −1/10.

The average rate of change is −1/10. The negative sign means the function decreased on average over the interval. The magnitude 1/10 tells you that for each unit increase in x, the output fell by 0.1. For f(x) = 1/x, this is the same as the slope of the secant line through (2, 0.5) and (5, 0.2).

The average rate of change always carries the units of f divided by the units of x. If f is distance in kilometres and x is time in hours, the result is km/h. If f is cost in dollars and x is items produced, the result is dollars per item. You must state the units in any applied problem; a bare number is incomplete.

Positive, Negative, and Zero Averages

• Positive: The output increased over the interval. The secant line slopes upward from left to right.
• Negative: The output decreased over the interval. The secant line slopes downward.
• Zero: The output started and ended at the same value. The secant line is horizontal. This can happen even if the function went up and back down in between.

Magnitude and Steepness

A larger absolute value means a steeper secant line. A very large positive number means the function rose rapidly. A value close to zero means the function changed little on average. But the average rate of change hides what happened inside the interval: a function could spike and then crash and still have a moderate average. The average is only about endpoints.

The average rate of change over an interval uses two distinct points and gives the slope of the secant line. The instantaneous rate of change at a single point uses the limit of the average rate as the interval shrinks to zero. That limit is the derivative f'(x).

The difference quotient (f(x + h) − f(x)) / h is the bridge. Keep h as a variable, simplify algebraically so that h cancels from the numerator, then let h → 0. The result is the instantaneous rate of change. For f(x) = x², the average rate of change from x to x + h is (2x + h). As h → 0, that becomes 2x, the derivative. OpenStax Calculus Volume 1, section 3.1 makes this progression explicit.

Stewart Calculus sections 2.7-2.8 treat the same idea: first find secant line slopes, then take the limit to find tangent line slopes. The derivative definition is f'(x) = lim_{h→0} (f(x + h) − f(x)) / h. Paul's Online Math Notes calls this "the definition of the derivative" and provides the step sequence for polynomial, rational, and radical functions.

When to Use Each

• Use the average rate of change when you need the overall trend over a specific interval, such as "what was the average speed between 2 pm and 4 pm?"
• Use the instantaneous rate of change when you need the rate at a precise moment, such as "what was the speed at exactly 3 pm?"

The most common error is treating the average rate of change as if it were the derivative. They are related but not the same. The average is a fixed number; the derivative is a limit. Always check whether the problem asks for the change "over [a, b]" or "at x = a".

Work these four problems. Each one uses a different representation. Answers are given after the set. If you get stuck, review the worked example for that representation.

1. From a table: A car's distance d(t) in metres at time t in seconds is recorded. t: 0, d: 0; t: 3, d: 45; t: 6, d: 120. Find the average speed over [0, 6] and over [3, 6].
2. From a graph: The graph of g(x) = −x² + 4x passes through (1, 3) and (3, 3). Find the average rate of change on [1, 3].
3. From a formula: Let h(x) = 2/(x − 1). Find the average rate of change on [2, 5].
4. Interpretation: The average rate of change of a stock's price over a week is −$2 per day. What happened to the price? Give the units of the answer.

Answers: (1) 20 m/s over [0, 6], 25 m/s over [3, 6]; (2) 0; (3) −1/2; (4) The stock price dropped on average $2 per day over the week. Units are dollars per day.