Why a Linear Function Has One Rate Everywhere
A straight line has the same steepness throughout its graph. In rate language, this means that the output changes by the same amount per unit of input no matter which interval you choose. The average-rate formula makes this observation precise and shows exactly where the rate comes from.
Here, a linear function means a function of the form \(f(x)=mx+b\), where \(m\) and \(b\) are fixed real numbers. The number \(m\) is the coefficient of the input, and \(b\) is the constant term. We will show that every average rate for this function is \(m\), provided the interval’s endpoints are distinct.
Average Rate on an Interval
For a function \(f\) and two distinct inputs \(x_1\) and \(x_2\), the average rate of change from \(x_1\) to \(x_2\) is the change in output divided by the change in input. The order of the endpoints must match in both differences.
This quotient compares two changes: \(f(x_2)-f(x_1)\), the output change, and \(x_2-x_1\), the input change. The endpoints must be distinct because otherwise the denominator is zero, so the average rate is not defined.
The same definition applies whether \(x_2\) is greater than or less than \(x_1\). If the direction is reversed, both changes reverse sign, leaving the quotient unchanged. For the linear function, we can see directly why every such quotient has the same value.
Deriving the Constant Average Rate
Let \(f(x)=mx+b\), and take any distinct inputs \(x_1\) and \(x_2\). Substitute the function values into the definition. The constant term \(b\) appears in both outputs, so it cancels when we subtract them.
Since \(x_1\ne x_2\), the quotient is defined.
Use \(f(x_1)=mx_1+b\) and \(f(x_2)=mx_2+b\).
The \(b\) terms cancel, and the remaining numerator factors as \(m(x_2-x_1)\).
Because \(x_2-x_1\ne 0\), dividing by it leaves \(m\).
This proves the claim for every pair of distinct inputs, not just for selected examples. The only condition needed for the calculation is \(x_1\ne x_2\). The coefficient \(m\) sets the rate, while \(b\) shifts every output by the same amount and therefore does not change output differences.
Worked Examples
Worked Example: An Increasing Linear Function
Let \(f(x)=3x-4\). Find the average rate of change from \(x_1=1\) to \(x_2=5\), then check the result.
First calculate both function values:
The output change is \(11-(-1)=12\), and the input change is \(5-1=4\). Therefore, the average rate is
Check: the input coefficient is \(m=3\), which agrees with the computed rate. Directly, \(3(5-1)=3(4)=12\), the output change we found. So the average rate on this interval is \(3\).
Worked Example: A Decreasing Linear Function
Let \(g(x)=-2x+5\). Find the average rate from \(x_1=-1\) to \(x_2=3\). This example also shows why a negative coefficient produces a negative rate.
Evaluate the function at both endpoints:
The output change is \(-1-7=-8\), and the input change is \(3-(-1)=4\). Thus
Check: the input coefficient is \(-2\), matching the rate. Also, multiplying the input change by that coefficient gives \((-2)(4)=-8\), exactly the output change. The function decreases by \(2\) output units for each \(1\) input unit as the input increases.
Worked Example: Reversing the Interval
Keep \(g(x)=-2x+5\), but calculate the rate from \(x_1=3\) to \(x_2=-1\). Now the input decreases, so both changes must be computed in this reversed order.
The endpoint values are already determined by substitution: \(g(3)=-1\) and \(g(-1)=7\). The output change, in the chosen order, is \(7-(-1)=8\), while the input change is \(-1-3=-4\). Therefore,
Check: the coefficient is still \(-2\). Reversing the interval changed the output change from \(-8\) to \(8\) and the input change from \(4\) to \(-4\); their quotient remains \(-2\). The order matters inside each difference, but reversing both differences does not alter the rate.
Worked Example: A Constant Linear Function
Let \(h(x)=6\), which can be written as \(h(x)=0x+6\). Find the average rate from \(x_1=2\) to \(x_2=7\).
Both function values are \(6\):
The output change is \(6-6=0\), and the input change is \(7-2=5\). Hence
Check: in \(h(x)=0x+6\), the input coefficient is \(m=0\), which agrees with the calculated rate. The function’s output does not change as the input changes.
The Coefficient and the Graph’s Slope
The graph of \(f(x)=mx+b\) is a straight line. The slope of a line through two distinct points is its vertical change divided by its horizontal change. On the graph of \(f\), the points corresponding to inputs \(x_1\) and \(x_2\) are \((x_1,f(x_1))\) and \((x_2,f(x_2))\). Their vertical change is \(f(x_2)-f(x_1)\), and their horizontal change is \(x_2-x_1\).
Thus the coefficient \(m\) is both the function’s average rate on every interval and the slope of its graph. If \(m>0\), the line rises as the input increases. If \(m<0\), it falls. If \(m=0\), it is horizontal. The constant \(b\) determines the output when \(x=0\), but does not affect the slope.
For example, in \(f(x)=3x-4\), the graph crosses the vertical axis at output \(-4\), while its slope is \(3\). These values have different roles: changing \(b\) moves the line up or down without changing its rate; changing \(m\) changes the rate and the line’s steepness.
A Common Mistake: Including the Intercept in the Rate
A frequent error is to treat the whole expression \(mx+b\) as the average rate. For instance, for \(f(x)=3x-4\), one might incorrectly claim that the rate is \(3-4=-1\). That calculation combines the input coefficient and constant term, even though they play different roles.
Check the claim on the interval from \(1\) to \(5\). The endpoint values are \(f(1)=-1\) and \(f(5)=11\), so the actual average rate is
not \(-1\). The correction is to calculate output change divided by input change, or apply the proved formula: for \(mx+b\), the rate is \(m\). The term \(b\) cancels in the output difference.
Another error is to reverse only one of the endpoint differences. From \(x_1=3\) to \(x_2=-1\) for \(g(x)=-2x+5\), using the reversed output change \(8\) but the unreversed input change \(4\) would give \(8/4=2\), the wrong sign. The matched changes are \(8\) and \(-4\), giving \(-2\). Keep the same endpoint order in numerator and denominator.
Why This Result Is Useful
For a linear function, one interval is enough to determine the average rate on every other interval: it must equal the input coefficient. Conversely, the coefficient can be interpreted from a graph as its rise or fall per unit of horizontal movement. This connects three descriptions of the same feature: the coefficient in the formula, the average rate calculated from values, and the slope of the straight-line graph.
The conclusion depends on the function having the form \(mx+b\). The cancellation in the proof comes from that specific form, so it should not be assumed for functions that are not linear. The essential test is not whether a function’s graph happens to be easy to inspect; it is whether its output differences simplify to \(m\) times the input difference.
Check Your Understanding
Use the definition and the result for linear functions to answer these questions.
- For \(p(x)=4x+7\), what is the average rate from \(x=0\) to \(x=6\)? Show the endpoint values and the quotient.
- For \(q(x)=-3x+2\), calculate the average rate from \(x=4\) to \(x=-2\). Why must both changes be written in the same order?
- Explain why changing \(b\) in \(f(x)=mx+b\) does not change the average rate.
- What is the slope and average rate of the graph of \(r(x)=9\)? Write this function in the form \(mx+b\).
- A student says the average rate of \(s(x)=5x-8\) is \(-3\), because \(5-8=-3\). Identify the mistake and give the correct rate.
