From Change in Values to a Graph Slope
A function assigns an output to each input in its domain. If we compare the outputs at two inputs, we can ask how much the output changed over the interval between them. That comparison is the function’s average rate of change on the interval.
The same comparison has a geometric interpretation. The two input-output pairs are points on the graph, and the line through them is called a secant line. Its slope measures the average change in output for each unit of input change between those endpoints.
This connection is useful because it gives one quantity two descriptions: an algebraic quotient of changes and a geometric slope. The quotient can be calculated directly from function values; the graph interpretation explains what that calculation represents.
Definition of Average Rate of Change
Let \(f\) be a function defined on an interval containing two distinct inputs \(a\) and \(b\). The input change from \(a\) to \(b\) is \(b-a\), and the corresponding output change is \(f(b)-f(a)\). The average rate of change is the output change divided by the input change.
When an interval is written as \([a,b]\) with \(a<b\), this is the average rate of change from the left endpoint \(a\) to the right endpoint \(b\). The numerator uses the output at the second endpoint minus the output at the first, and the denominator uses the second input minus the first. Keeping the order consistent is essential.
The quotient compares the total output change with the total input change. It is an average over the interval, not a claim that the output changes by this amount at every point in the interval. Nor does calculating it require the function to be continuous: it is enough that the function is defined at both endpoints. If the function is defined throughout the interval, that is more than is needed for this endpoint calculation.
Why the Quotient Is a Secant Slope
The graph of \(f\) consists of points whose coordinates have the form \((x,f(x))\). At the two selected inputs, the endpoint points are therefore \((a,f(a))\) and \((b,f(b))\). Since \(a\ne b\), these are distinct points: their first coordinates differ.
The slope of a nonvertical line through two points is the difference in their vertical coordinates divided by the difference in their horizontal coordinates, using the same order for both differences. Applied to these endpoint points, that definition gives the following derivation.
The graph points are \((a,f(a))\) and \((b,f(b))\).
For points \((x_1,y_1)\) and \((x_2,y_2)\), the slope is \((y_2-y_1)/(x_2-x_1)\), provided \(x_2\ne x_1\).
Here \(x_1=a\), \(y_1=f(a)\), \(x_2=b\), and \(y_2=f(b)\). Thus the slope is \((f(b)-f(a))/(b-a)\).
This calculation proves that the average rate of change and the slope of the secant through the two graph points are the same number. The result follows directly from the definition of slope; no extra condition such as continuity is needed for the endpoint secant.
The line through those two graph points is the secant line for the chosen pair of inputs. It is a line determined by two points, so its slope describes the overall change between the endpoints. It does not, by itself, describe the function’s behavior at every input between them.
Worked Examples
Worked Example: A Reciprocal Function
Find the average rate of change of \(f(x)=1/x\) on \([1,2]\), then verify that it is the secant slope.
First calculate the endpoint outputs:
The output change from the first endpoint to the second is \(f(2)-f(1)\), and the input change is \(2-1\). Therefore,
The endpoint graph points are \((1,1)\) and \((2,\frac12)\). Their slope is
The two computations agree, as the secant-slope result predicts. As a direct check, the line through \((1,1)\) with slope \(-\frac12\) is \(y-1=-\frac12(x-1)\). At \(x=2\), this gives \(y-1=-\frac12\), so \(y=\frac12\), the correct second endpoint value.
Worked Example: A Square-Root Function
Find the average rate of change of \(g(x)=\sqrt{x}\) on \([4,9]\), and connect it to the secant through the endpoint points.
Evaluate the function at each endpoint:
The input change is \(9-4=5\), and the output change is \(3-2=1\). Thus,
The endpoint graph points are \((4,2)\) and \((9,3)\). Their secant slope is
To check the line, use the point \((4,2)\) and slope \(\frac15\): \(y-2=\frac15(x-4)\). Substituting \(x=9\) gives \(y-2=\frac15(5)=1\), hence \(y=3\), which matches \(g(9)\). The average rate and secant slope are both \(\frac15\).
Worked Example: A Cubic Function
For \(h(x)=x^3\), calculate the average rate of change on \([1,2]\), then check it using the endpoint secant.
Compute the endpoint values:
The output change is \(8-1=7\), while the input change is \(2-1=1\). Therefore,
The points on the graph are \((1,1)\) and \((2,8)\), so their slope is
For a further check, the line through \((1,1)\) with slope \(7\) is \(y-1=7(x-1)\). At \(x=2\), it gives \(y-1=7\), so \(y=8\), exactly the other endpoint value. The calculation and the secant interpretation agree.
Interpreting the Sign When \(a<b\)
For an interval written with \(a<b\), the denominator \(b-a\) is positive. Consequently, the sign of the average rate of change is determined by the output change \(f(b)-f(a)\).
- If the average rate is positive, then \(f(b)>f(a)\): the output at the second endpoint is greater than the output at the first.
- If the average rate is negative, then \(f(b)<f(a)\): the output at the second endpoint is smaller than the output at the first.
- If the average rate is zero, then \(f(b)=f(a)\): the endpoint outputs are equal.
These conclusions compare the two endpoint values only. For instance, a positive average rate from \(a\) to \(b\) does not establish that every output between the endpoints is larger than the preceding one. The quotient records the net endpoint change divided by the input interval length.
A Common Mistake: Forgetting to Divide by the Input Change
A frequent error is to calculate the output difference and report it as the average rate. The output difference is only the numerator; the rate also depends on how large the input interval is.
For example, let \(f(x)=x^2\) on \([1,3]\). The endpoint values are \(f(1)=1\) and \(f(3)=9\), so the output change is \(9-1=8\). Reporting \(8\) as the average rate is incorrect because the input change is \(3-1=2\), not \(1\). The correct calculation is
The secant points are \((1,1)\) and \((3,9)\), whose slope is also \((9-1)/(3-1)=8/2=4\). The answer \(8\) is the total output change; \(4\) is the average output change per unit of input change over this interval.
What the Secant Connection Gives Us
The secant interpretation makes the formula easier to remember and helps check a calculation. Identify the two graph points, take the vertical difference in a consistent order, and divide by the corresponding horizontal difference in that same order. The result is the slope of the line through the endpoints.
It also clarifies the scope of the quantity. The secant slope is determined by the selected pair of inputs. Choosing different endpoints generally gives a different pair of points and therefore may give a different average rate. A single secant slope summarizes the change across its particular interval; it does not describe all intervals at once.
Most importantly, the geometric interpretation is not a separate rule to memorize. It is exactly the slope formula applied to the graph’s endpoint coordinates. Whenever the function values at distinct inputs are known, the same quotient both calculates the average rate of change and gives the secant slope.
Check Your Understanding
Use the definition and the endpoint secant interpretation to answer the following questions.
- State the formula for the average rate of change of \(f\) from \(a\) to \(b\), and explain why \(a\) and \(b\) must be distinct.
- What are the coordinates of the two graph points used to form the secant for inputs \(a\) and \(b\)?
- For \(f(x)=x+2\) on \([1,4]\), calculate the average rate of change and verify it as a secant slope.
- On an interval with \(a<b\), what does a negative average rate tell you about \(f(b)\) and \(f(a)\)?
- For \(f(x)=x^2\) on \([2,5]\), someone reports the output difference \(25-4=21\) as the average rate. What input change did they omit, and what is the correct average rate?
