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Derivative as a Limit · Tutorial 81 of 999

Average Rate of Change as Secant Slope

Calculate average rate of change from endpoint values and interpret the result as the slope of a secant line.

Beginner 10 min read

What You'll Learn

  • Calculate average rate of change using the two endpoint values of a function.
  • Identify the points on a graph that determine the corresponding secant line.
  • Explain why average rate of change equals secant slope.
  • Interpret positive, negative, and zero average rates.
  • Include units when interpreting rates in applied contexts.
  • Avoid confusing a change in function value with an average rate of change.

From Change to Rate

A function can describe how one quantity depends on another. To measure how much the output changes between two inputs, subtract the function values. To measure the change per unit of input, divide that output change by the input change.

This quotient is the function’s average rate of change over an interval. On a graph, it has a geometric interpretation: it is the slope of the straight line through the two points corresponding to the interval’s endpoints.

Definition: Let \(f\) be a function, and let \(a\) and \(b\) be distinct inputs in its domain. The average rate of change of \(f\) from \(x=a\) to \(x=b\) is
$$ \frac{f(b)-f(a)}{b-a}. $$
The numerator is the change in output, and the denominator is the change in input.

When describing an interval from left to right, we usually take \(a<b\). The definition also works if the endpoints are listed in reverse order: reversing them changes the signs of both differences, so the quotient is unchanged.

The Geometric Meaning: Secant Slope

The graph of \(f\) contains the points \((a,f(a))\) and \((b,f(b))\). The straight line through these two distinct points is called a secant line to the graph over the interval from \(a\) to \(b\).

The slope of a nonvertical line through \((x_1,y_1)\) and \((x_2,y_2)\), where \(x_1\ne x_2\), is the change in vertical coordinate divided by the change in horizontal coordinate. Applying that definition to the two points on the graph gives

$$ \text{secant slope} = \frac{f(b)-f(a)}{b-a} = \text{average rate of change}. $$

This equality is not a coincidence. Both quantities compare the same vertical change, \(f(b)-f(a)\), with the same horizontal change, \(b-a\). The rate is the numerical change-per-input; the secant slope is the geometric description of that same quotient.

1
Find the endpoint inputs.
Use the two distinct values \(a\) and \(b\) that bound the interval.
2
Evaluate the function at both endpoints.
Find \(f(a)\) and \(f(b)\), which give the vertical coordinates of the graph points.
3
Compute output change divided by input change.
Calculate \(\frac{f(b)-f(a)}{b-a}\). This is both the average rate and the secant slope.
4
Interpret the result.
Use its sign and, when appropriate, its units to explain what the rate or slope means.

Worked Examples

Worked Example: A Quadratic Function on an Interval

Find the average rate of change of \(f(x)=x^2+1\) from \(x=1\) to \(x=4\), then find the equation of the secant line.

First evaluate the function at each endpoint:

$$ f(1)=1^2+1=2, \qquad f(4)=4^2+1=17. $$

The two graph points are \((1,2)\) and \((4,17)\). The average rate of change is

$$ \frac{f(4)-f(1)}{4-1} = \frac{17-2}{3} = \frac{15}{3} = 5. $$

Thus, the secant slope is \(5\). Using the point-slope form of a line through \((1,2)\), the secant line is \(y-2=5(x-1)\), or \(y=5x-3\).

Check: At \(x=1\), the line gives \(y=5(1)-3=2\). At \(x=4\), it gives \(y=5(4)-3=17\). It passes through both required points, and its slope is \(5\), matching the calculated average rate.

Worked Example: Interpreting a Negative Rate with Units

A moving object’s position, in meters, is described by \(s(t)=20+3t-t^2\), where \(t\) is measured in seconds. Find its average rate of change from \(t=1\) second to \(t=4\) seconds.

Evaluate the position at the two times:

$$ s(1)=20+3(1)-1^2=20+3-1=22\text{ m}, $$ $$ s(4)=20+3(4)-4^2=20+12-16=16\text{ m}. $$

The time change is \(4-1=3\) seconds, and the position change is \(16-22=-6\) meters. Therefore,

$$ \frac{s(4)-s(1)}{4-1} = \frac{16-22}{3} = \frac{-6}{3} = -2\text{ m/s}. $$

The negative sign means that the position at the later endpoint is less than the position at the earlier endpoint. The average rate over this interval is \(-2\) meters per second.

Check: The endpoint positions differ by \(16-22=-6\) meters. At \(-2\) meters per second for \(3\) seconds, the corresponding change is \((-2)(3)=-6\) meters, as required.

Worked Example: A Reciprocal Function

Find the average rate of change of \(g(x)=\frac{1}{x}\) from \(x=1\) to \(x=3\), and verify the slope using the secant line.

Evaluate the function at the endpoints:

$$ g(1)=\frac{1}{1}=1, \qquad g(3)=\frac{1}{3}. $$

The graph points are \((1,1)\) and \((3,\frac{1}{3})\). Their secant slope, and thus the average rate of change, is

$$ \frac{g(3)-g(1)}{3-1} = \frac{\frac{1}{3}-1}{2} = \frac{-\frac{2}{3}}{2} = -\frac{1}{3}. $$

Using \((1,1)\), an equation for the secant line is \(y-1=-\frac{1}{3}(x-1)\). Simplifying gives \(y=-\frac{x}{3}+\frac{4}{3}\).

Check: At \(x=1\), this line gives \(-\frac{1}{3}+\frac{4}{3}=1\). At \(x=3\), it gives \(-1+\frac{4}{3}=\frac{1}{3}\). Its slope is \(-\frac{1}{3}\), agreeing with the quotient.

Why the Formula Gives the Secant Slope

The connection can be justified directly from the definition of slope. The secant line joins the graph points \(P=(a,f(a))\) and \(Q=(b,f(b))\). Because \(a\ne b\), the horizontal change between them is nonzero, so the line is not vertical and its slope is defined.

1
Write the coordinates.
The first point is \(P=(a,f(a))\), and the second is \(Q=(b,f(b))\).
2
Apply the slope definition.
Slope is vertical change divided by horizontal change, so the secant slope is \(\frac{f(b)-f(a)}{b-a}\).
3
Compare with average rate.
By definition, average rate of change is also \(\frac{f(b)-f(a)}{b-a}\). Since the expressions are identical, the average rate equals the secant slope.

This argument requires only that \(a\) and \(b\) be distinct inputs where \(f\) is defined. It does not require the function to be continuous at every point between the endpoints. The two endpoint values are enough to calculate the quotient and define the line through the corresponding graph points.

Reading the Sign and Size

The sign of the average rate describes the direction of the endpoint-to-endpoint change. If \(b>a\), then the denominator \(b-a\) is positive. In that case:

  • If \(f(b)>f(a)\), the average rate is positive, and the secant line rises from left to right.
  • If \(f(b)<f(a)\), the average rate is negative, and the secant line falls from left to right.
  • If \(f(b)=f(a)\), the average rate is zero, and the secant line is horizontal.

The size of the rate measures the output change per unit of input over the whole interval. A larger absolute value means a steeper secant line. It does not say that the function changes at that same rate at every point inside the interval; it summarizes the net change between the endpoints.

Units follow the same division. If output is measured in meters and input in seconds, average rate has units of meters per second. If both quantities are pure numbers, the rate has no physical units. Keeping units in the quotient is a useful way to check whether the calculation has been set up correctly.

Common Mistake: Dividing by the Wrong Quantity

A frequent error is to average the two function values instead of dividing their difference by the input difference. For \(f(x)=x^2\) from \(x=1\) to \(x=3\), a mistaken calculation might be

$$ \frac{f(1)+f(3)}{2} = \frac{1+9}{2} = 5. $$

This is the average of the endpoint outputs, not the average rate of change. The rate requires change in output divided by change in input:

$$ \frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = \frac{8}{2} = 4. $$

Check: The secant line through \((1,1)\) and \((3,9)\) has slope \(\frac{9-1}{3-1}=4\), not \(5\). The wrong calculation used the endpoint values but did not measure either the output change or the input change.

What Average Rate Does—and Does Not—Tell You

Average rate of change condenses the behavior over an interval into one number. It is useful for comparing net changes, describing how an output responds on average to an input change, and finding the slope of the line joining two points on a graph.

It does not, by itself, describe what happens between the endpoints. For example, equal endpoint values give an average rate of zero, even if the function takes other values between them. Zero average rate means only that the net output change between the chosen endpoints is zero.

Likewise, an average rate is not automatically an instantaneous rate at either endpoint or at some particular input. Here, the calculation uses two distinct inputs and their secant line. The central task is to identify those endpoints, evaluate the function correctly, and divide the change in output by the change in input.

Check Your Understanding

For each question, show the endpoint values and the change in input as well as the final quotient.

  1. For \(f(x)=3x+2\), find the average rate of change from \(x=-1\) to \(x=2\). What is the slope of the corresponding secant line?
  2. For \(g(x)=x^2-1\), calculate the average rate of change from \(x=0\) to \(x=3\). Identify the two graph points used.
  3. A quantity \(p(t)\), measured in liters, changes from \(14\) liters at \(t=2\) minutes to \(8\) liters at \(t=5\) minutes. What is its average rate of change, including units?
  4. If \(h(a)=h(b)\) for distinct inputs \(a\) and \(b\), what is the average rate of change from \(a\) to \(b\)? Describe the secant line.
  5. Explain why reversing the order of the same two endpoints does not change the average rate of change.

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