Rates at a Breakpoint
A piecewise model uses different formulas in different ranges. In a tiered bill or tax schedule, the formula may change when usage or income reaches a specified threshold. At that threshold, the important question is not simply which formula applies on either side. We must ask what rate the model has at the breakpoint itself.
The rates approaching from below and above may differ. If they do, the model has no single derivative there, even when its value is continuous. Alternatively, the formulas may change while the two rates agree, in which case a derivative can exist. If the model jumps in value, the difference quotients can fail to have finite limits at all.
One-Sided Derivatives
Suppose \(f\) is defined at \(c\) and on intervals immediately to its left and right. For \(h\ne 0\), the difference quotient at \(c\) is \(\frac{f(c+h)-f(c)}{h}\). Its limit as \(h\) approaches zero through negative values is the left-hand derivative; its limit as \(h\) approaches zero through positive values is the right-hand derivative, provided the respective limits exist as finite numbers.
The ordinary derivative \(f'(c)\) exists exactly when these two one-sided derivatives both exist and are equal. If they differ, the model has different limiting rates from the two sides, so there is no single derivative at \(c\). If either one-sided limit is not finite or does not exist, there is also no derivative at \(c\).
For a continuous piecewise-linear model, each one-sided derivative is the slope of the corresponding line segment. Continuity matters: the difference quotient uses \(f(c)\), and if one side of the model does not meet that value, the quotient includes the effect of the jump as well as the slope of the branch.
Worked Example: A Billing Rate Changes at a Usage Threshold
Worked Example: A Billing Rate Changes at a Usage Threshold
A monthly bill \(B(x)\), measured in dollars, depends on usage \(x\), measured in units. The first 100 units cost $4 per unit; each additional unit costs $6. The model is
At the breakpoint, \(B(100)=4(100)=400\) dollars. For a negative \(h\) sufficiently close to zero, \(100+h\le 100\), so
Thus \(B'_{-}(100)=4\) dollars per unit. For positive \(h\), the second branch applies:
Thus \(B'_{+}(100)=6\) dollars per unit. The bill is continuous at 100 units, but the one-sided rates differ. Therefore \(B'(100)\) does not exist: the marginal rate changes abruptly from $4 per unit to $6 per unit at the threshold.
Check: The left branch gives \(B(100)=400\), and the right branch approaches \(400+6(0)=400\) as usage decreases to 100 from above. The computed one-sided quotients are exactly the slopes of the two continuous line segments.
Worked Example: Different Formulas, One Rate at the Threshold
Worked Example: Different Formulas, One Rate at the Threshold
Consider a tax model \(T(y)\), measured in dollars, where \(y\) is income in dollars. Up to $20,000, tax is 10% of income. Above $20,000, the model adds a small smoothly increasing surcharge to the 10% tax amount:
At \(c=20000\), the value from the first branch is \(T(20000)=0.10(20000)=2000\) dollars. Write \(h\) for the change in income. For \(h<0\),
For \(h>0\), substitute into the second branch:
As \(h\to 0^+\), this quotient approaches \(0.10\). Hence \(T'_{-}(20000)=T'_{+}(20000)=0.10\), and \(T'(20000)=0.10\) dollars of tax per dollar of income. The derivative exists even though the formulas on the two sides are different. The additional quadratic term contributes zero to the limiting rate exactly at the threshold.
Check: At \(y=20000\), the left formula gives $2,000. The right formula approaches \(2000+0+0=2000\) dollars. The left quotient is \(0.10\), and the right quotient tends to \(0.10\), so both the values and the limiting rates agree at the breakpoint.
Why Matching One-Sided Rates Is the Right Test
The derivative is defined by a two-sided limit of the difference quotient. A two-sided limit exists when the quotient approaches the same finite value from the left and from the right. The following statement makes that test precise.
Proof. If \(f'(c)\) exists, then the difference quotient has a limit as \(h\to 0\). Restricting \(h\) to negative values or to positive values cannot change that limit. Therefore both one-sided derivatives exist and equal \(f'(c)\), so they equal each other.
Conversely, suppose both one-sided derivatives exist and equal the same finite number \(L\). Let \(\varepsilon>0\). From the left-hand limit, there is a positive \(\delta_-\) such that when \(-\delta_-<h<0\), the quotient differs from \(L\) by less than \(\varepsilon\). From the right-hand limit, there is a positive \(\delta_+\) such that when \(0<h<\delta_+\), the quotient differs from \(L\) by less than \(\varepsilon\). Set \(\delta=\min(\delta_-,\delta_+)\). Whenever \(0<|h|<\delta\), either \(h<0\) or \(h>0\), and the corresponding one-sided condition gives
This is the definition of the two-sided limit of the difference quotient being \(L\). Thus \(f'(c)=L\). This proves the result.
Worked Example: A Fixed Charge Creates a Jump
Worked Example: A Fixed Charge Creates a Jump
A service bill \(F(q)\), measured in dollars, has a fixed $30 charge once usage exceeds 50 units:
The model assigns \(F(50)=8(50)=400\) dollars. For \(h<0\), the left-hand quotient is
For \(h>0\), the second branch gives \(F(50+h)=400+10h+30=430+10h\). Therefore
As \(h\to0^+\), \(30/h+10\) grows without bound, so there is no finite right-hand derivative. Consequently \(F'(50)\) does not exist. Although the line used above the threshold has slope 10 dollars per unit, that slope is not the right-hand derivative at the breakpoint: the $30 jump also affects the difference quotient.
Check: The left-hand value at the breakpoint is $400, while the right-hand formula approaches \(400+0+30=430\) dollars. The $30 difference is exactly the term \(30/h\) in the right-hand quotient, confirming that the model jumps and has no finite right-hand rate there.
A Common Mistake: Averaging the Two Rates
When the one-sided rates differ, averaging them does not produce a derivative. In the billing example with a $4-per-unit rate below 100 units and a $6-per-unit rate above it, one might calculate
and incorrectly report a derivative of $5 per unit at 100 units. But the difference quotient is exactly 4 for every sufficiently small negative \(h\), and exactly 6 for every sufficiently small positive \(h\). It does not approach 5 from either side. The correction is to compare the two one-sided limits: because \(4\ne6\), there is no derivative at the breakpoint.
It is also important not to confuse a branch’s slope with the derivative at the breakpoint when the branch values do not meet. In the fixed-charge example, the upper branch has slope 10, but its difference quotient at the breakpoint includes \(30/h\). The actual right-hand rate at the breakpoint is therefore not 10.
Interpreting a Breakpoint Carefully
At a threshold in a tiered model, distinguish three possibilities:
- Continuous model, unequal one-sided rates: the value joins, but the rate changes abruptly; there is no derivative at the breakpoint.
- Continuous model, equal one-sided rates: a derivative exists, even if the formulas on either side differ.
- Jump in the model’s value: the difference quotient includes the jump, so a finite one-sided derivative may fail to exist.
For billing and tax models, the derivative describes a local change in the modeled amount per additional unit of usage or income. At a breakpoint, compute the one-sided difference quotients from the model’s assigned value at that point. This determines whether a single instantaneous rate is meaningful there, rather than merely identifying the slopes of the neighboring formulas.
Check Your Understanding
Use one-sided difference quotients where appropriate.
- A continuous piecewise-linear cost model has slope 7 below a breakpoint and slope 9 above it. Does its derivative exist at the breakpoint? Explain.
- What condition on the two one-sided derivatives guarantees that the ordinary derivative exists?
- Why can two different formulas on either side of a breakpoint still produce a derivative there?
- A model has a fixed charge that makes its value jump at a threshold. Why might the slope of the upper branch fail to be the right-hand derivative at the threshold?
- For the tax model in the worked example, what is the derivative at income $20,000, in units of tax dollars per income dollar?
