How Input Scaling Changes Average Rates

Why Input Units Affect a Numerical Rate

Suppose a quantity changes as an input changes. An average rate describes the output change divided by the input change. If the same input is recorded using a different scale, the output change may stay the same while the numerical input change changes. The numerical rate must then change as well.

For example, an elapsed time of one hour is also sixty minutes. A change in position over that hour has the same size whichever time unit is used, but dividing it by one hour or by sixty minutes gives different numerical rates. The units explain how those rates are related.

We can express this relationship for a general function. The key is to match the intervals: if the new input is multiplied by a constant before it is fed into the original function, then the endpoints of the interval must be scaled by that same constant.

Average Rate and the Scaled Function

Now let \(c\) be a constant and define a new function \(g\) by \(g(x)=f(cx)\), wherever \(cx\) belongs to the domain of \(f\). The input \(x\) of \(g\) is first scaled to \(cx\), and then \(f\) is evaluated there.

For distinct values \(a\) and \(b\), the average rate of \(g\) from \(a\) to \(b\) is \(\frac{g(b)-g(a)}{b-a}\). When \(c\ne 0\), the corresponding inputs for \(f\) are \(ca\) and \(cb\), which are also distinct. Comparing these average rates gives the scaling rule.

Why the Scaling Formula Works

The result follows from substituting the definition of \(g\) and rewriting the input difference. The denominator for the corresponding interval of \(f\) is \(cb-ca=c(b-a)\). Because \(c\ne 0\) and \(a\ne b\), neither denominator is zero.

In equations, the complete calculation is

This argument proves the rule directly from the definition of average rate. It does not require \(f\) to be linear or differentiable. The only requirements are that the function values exist and that the two inputs in each quotient are distinct.

Worked Example: Scaling an Input by a Positive Factor

What the Sign of the Scale Factor Means

If \(c>0\), the matched inputs \(ca\) and \(cb\) remain in the same order as \(a\) and \(b\). The numerical average rate is multiplied by a positive factor, so its sign is unchanged.

If \(c<0\), the matched inputs reverse order. The average-rate quotient is still defined: reversing the order of its endpoints changes the signs of both differences, leaving the quotient unchanged. The theorem nevertheless multiplies that quotient by the negative number \(c\), so the average rate of \(g\) has the opposite sign from the corresponding average rate of \(f\) whenever that rate is nonzero; if it is zero, the rate of \(g\) is also zero.

The Zero Scale Is a Special Case

The scaling formula requires \(c\ne 0\). If \(c=0\), then \(g(x)=f(0)\) for every allowed \(x\). Provided \(0\) is in the domain of \(f\), the function \(g\) is constant, so for distinct \(a\) and \(b\) its average rate is

In this case, the two inputs \(ca\) and \(cb\) both equal zero. They do not form a valid interval for an average-rate quotient, because their difference is zero. The scaled-endpoint formula therefore cannot be used: the corresponding interval has collapsed to a single input.

Worked Example: Converting an Hourly Rate to a Per-Minute Rate

Rates, Scales, and Units

The formula concerns numerical input values. When those values represent measurements, the units clarify which scale factor applies. If \(x\) is the numerical input in one unit and \(z\) is its numerical value in another unit, write the relationship explicitly before comparing rates.

For example, in the preceding conversion, the old input was time in hours and the new input was time in minutes. One minute represents \(1/60\) of an hour, so the function expressed in minutes uses the old input \(m/60\). Its average rate per minute is \(1/60\) of the corresponding rate per hour. Conversely, converting a per-minute rate to a per-hour rate multiplies its numerical value by \(60\).

Keep the output units fixed while making this comparison. A rate such as metres per hour and a rate such as metres per minute describe the same kind of change, but their numerical values differ because the input units differ. Dimensional analysis helps keep this relationship clear.

A Common Mistake

A frequent error is to use the average output change for the scaled interval but divide by the unscaled input change, then claim the result is the original average rate. For example, in the first worked example the output change is \(36-9=27\). The corresponding input change for \(f\) is \(6-3=3\), while the input change for \(g\) is \(2-1=1\). Using the wrong denominator gives

The value \(9\) is the average rate of \(f\) across inputs \(3\) to \(6\); it is not the average rate of \(g\) across inputs \(1\) to \(2\). The output change is shared, but each rate must be divided by the input change for its own interval. In general, using the input change \(b-a\) instead of \(c(b-a)\) therefore multiplies the quotient by \(c\).

Another useful check is to ask whether the transformed endpoints are correct. When \(g(x)=f(cx)\), the inputs to \(f\) are \(ca\) and \(cb\), not \(a\) and \(b\). Omitting this endpoint match can lead to comparing rates across different intervals.