From an Average Rate to an Increment
An average rate compares the change in a function’s output with the change in its input over two distinct inputs. When those inputs are named \(x\) and \(x+h\), the change in input has a particularly simple form: it is \(h\). This lets us write the average rate in a compact and reusable way.
The number \(h\) is called an increment because it tells us how much the input changes from the starting value \(x\). It may be positive or negative. The two inputs being compared are \(x\) and \(x+h\), and the corresponding graph points are \((x,f(x))\) and \((x+h,f(x+h))\).
The name “difference quotient” describes its structure: it is a quotient, or ratio, of two differences. The numerator \(f(x+h)-f(x)\) is the change in output. The denominator \(h\) is the corresponding change in input. Neither difference can be omitted without changing what is being measured.
Why the Denominator Is \(h\)
For any two distinct inputs \(a\) and \(b\) in the domain of \(f\), the average rate from \(a\) to \(b\) is the output change divided by the input change. In symbols, it is \(\frac{f(b)-f(a)}{b-a}\). To use this definition for the inputs \(x\) and \(x+h\), substitute \(a=x\) and \(b=x+h\). The input difference is
The output difference is \(f(b)-f(a)=f(x+h)-f(x)\). Substituting both differences into the definition gives
This establishes the formula directly: the difference quotient is not a new kind of rate, but the average rate written for inputs separated by an increment. The denominator is \(h\) precisely because \(h\) is the distance in input values, with its sign retained, from \(x\) to \(x+h\).
Reading the Secant Endpoints
On the graph of \(y=f(x)\), the difference quotient compares the two points
The straight line through two distinct points on a graph is a secant line. Its slope is the change in vertical coordinate divided by the change in horizontal coordinate. For these endpoints, that slope is
This graph interpretation gives a useful check on the formula: each function value in the numerator must match the input named in the same endpoint. The term \(f(x+h)\) belongs to the point with horizontal coordinate \(x+h\), while \(f(x)\) belongs to the point with horizontal coordinate \(x\).
Worked Examples
Worked Example: A Linear Function Over an Increment
Let \(f(t)=2t+1\), take \(x=3\), and use the increment \(h=2\). Find the difference quotient and identify the secant endpoints.
First check the inputs: \(x=3\) and \(x+h=3+2=5\). Both are valid inputs for this linear function, and \(h=2\ne 0\).
The output change is \(f(5)-f(3)=11-7=4\), and the input change is \(h=2\). Thus
The secant endpoints are \((3,7)\) and \((5,11)\). Checking from the coordinates gives slope \(\frac{11-7}{5-3}=\frac{4}{2}=2\), which agrees with the difference quotient.
Worked Example: A Square-Root Function
Let \(f(t)=\sqrt{t}\), with \(x=4\) and \(h=5\). Calculate the average rate from \(x\) to \(x+h\), and state the secant endpoints.
The inputs are \(4\) and \(4+5=9\). Both are in the domain of the square-root function, and \(5\ne 0\). Evaluate the function at each input:
The output change is \(3-2=1\). The input change is \(9-4=5\), which is also the increment \(h\). Therefore
The graph points are \((4,2)\) and \((9,3)\). Their coordinate slope is \(\frac{3-2}{9-4}=\frac{1}{5}\). This confirms that the difference quotient gives the secant slope between the stated endpoints.
Worked Example: A Negative Increment
Let \(f(t)=\frac{1}{t}\), take \(x=2\), and choose \(h=-\frac{1}{2}\). Find the difference quotient and interpret its endpoints.
The second input is
Neither input is zero, so both \(2\) and \(\frac{3}{2}\) are in the domain of \(f\). Also, \(h=-\frac{1}{2}\ne 0\). The function values are
The output change from \(x\) to \(x+h\) is
Divide that change by the input change \(h=-\frac{1}{2}\):
The endpoints in the order from \(x\) to \(x+h\) are \((2,\frac{1}{2})\) and \((\frac{3}{2},\frac{2}{3})\). Their slope, using that same order, is \(\frac{\frac{2}{3}-\frac{1}{2}}{\frac{3}{2}-2} =\frac{\frac{1}{6}}{-\frac{1}{2}} =-\frac{1}{3}\). If the endpoints are reversed, both coordinate differences change sign, so their ratio remains \(-\frac{1}{3}\). The negative increment changes the direction of the input change; it does not invalidate the quotient.
Using the Formula Reliably
When a problem gives \(f\), \(x\), and \(h\), keep the roles of the quantities clear. The following sequence makes those roles explicit.
Confirm \(h\ne 0\), and confirm that both \(x\) and \(x+h\) belong to the domain of \(f\).
Calculate \(x+h\) before evaluating the function there. This value is the second horizontal coordinate.
Find \(f(x+h)\) and \(f(x)\), keeping each value paired with its input.
Subtract the starting output from the ending output, then divide by \(h=(x+h)-x\). Check the result against the slope between the two graph points.
The order in the numerator matters: \(f(x+h)-f(x)\) is the output at the second input minus the output at the first. The denominator uses the matching order, \((x+h)-x\). Reversing both differences gives the same slope, but reversing only one changes its sign.
A Common Mistake: Leaving Out the Starting Output
A frequent error is to divide \(f(x+h)\) by \(h\), forgetting that the numerator must be the change in output. For example, with \(f(t)=2t+1\), \(x=3\), and \(h=2\), the incorrect calculation would be
This is not the average rate between inputs \(3\) and \(5\): it uses the output at \(5\) but does not account for the starting output at \(3\). The correct output change is \(f(5)-f(3)=11-7=4\), so the correct quotient is
The endpoint check catches the error: the points are \((3,7)\) and \((5,11)\), and their slope is \(\frac{11-7}{5-3}=2\), not \(\frac{11}{2}\).
What the Increment Does—and Does Not—Say
The increment \(h\) specifies a second input relative to a chosen starting input \(x\). A positive \(h\) makes \(x+h\) larger than \(x\); a negative \(h\) makes it smaller. In either case, the average rate uses the output difference and the corresponding input difference. The quotient is meaningful only when both inputs are allowed for the function and are distinct.
The formula itself does not require \(f\) to be linear, continuous, or differentiable. It only requires that the two function values exist and that the inputs differ. For a particular function and pair of inputs, the quotient gives one average rate across that interval. It should not be confused with a rate at a single input: it describes the change between two endpoints.
This notation is useful because it makes the two endpoints visible while expressing their separation with one number, \(h\). It also provides a consistent way to describe secant slopes for many different functions and intervals. The essential check remains the same: the numerator must be the output change between those endpoints, and the denominator must be their input change.
Check Your Understanding
For each question, keep track of the starting input, the increment, and the second endpoint.
- For \(f(t)=t+4\), \(x=1\), and \(h=3\), identify both secant endpoints and calculate the difference quotient.
- In \(\frac{f(x+h)-f(x)}{h}\), what change is represented by the numerator, and what change is represented by the denominator?
- Why must \(h\ne 0\), and what other condition must \(x+h\) satisfy?
- If \(h<0\), does the formula stop describing a secant slope? Explain how the input change is represented.
- For \(f(t)=\frac{1}{t}\), \(x=1\), and \(h=-2\), are both required inputs in the domain? Explain before attempting to form the quotient.
