Why a Target Input Can Be Unique
An accumulation function records how a rate adds to or subtracts from a starting amount. If the rate keeps accumulation moving in one direction, then the same accumulated target cannot be reached at two different inputs. This observation is useful when reading rate graphs: their signed areas describe the changes in accumulation, while their signs can help establish whether a target input is unique.
The word strictly is important. A function may never move in the opposite direction and still remain constant over an interval. In that case, a target value on the flat portion is reached at many inputs. The uniqueness result therefore requires strict monotonicity, not merely a general tendency to increase or decrease.
Accumulation and Strict Monotonicity
An accumulation function based at an input \(a\) has the form
where \(r(t)\) is the rate. The signed integral measures the net change in accumulation between \(a\) and \(x\). In particular, for \(x<y\), additivity of the definite integral gives
This relation is the key connection between a rate graph and the behavior of its accumulation function. A positive signed area from \(x\) to \(y\) means accumulation increases over that interval; a negative signed area means it decreases.
Why Strict Monotonicity Gives Uniqueness
Suppose \(A\) is strictly increasing on \(I\). If two different inputs \(x_1\) and \(x_2\) both produced the target \(T\), then one of them would be smaller. Write \(x_1<x_2\). Strict increase would require \(A(x_1)<A(x_2)\), which is incompatible with both values equalling \(T\). For a strictly decreasing function, the required inequality is reversed, and the same contradiction follows.
Suppose \(A(x_1)=T\) and \(A(x_2)=T\), with \(x_1\ne x_2\).
Since they are different, either \(x_1<x_2\) or \(x_2<x_1\). Relabel them so that \(x_1<x_2\).
If \(A\) is strictly increasing, then \(A(x_1)<A(x_2)\); if it is strictly decreasing, then \(A(x_1)>A(x_2)\).
The assumption that both equal \(T\) gives \(A(x_1)=A(x_2)\), contradicting either strict inequality. Thus two distinct target inputs cannot exist.
This proof establishes “at most one,” not “exactly one.” Strict monotonicity rules out a repeated target value, but the function might not reach the chosen target anywhere in its domain. Establishing existence is a separate question.
Reading Strict Monotonicity From a Rate Graph
For an accumulation function, compare any two inputs \(x<y\) using the signed-area relation \(A(y)-A(x)=\int_x^y r(t)\,dt\). If this integral is positive for every pair \(x<y\) in the domain, then \(A(y)>A(x)\) for every such pair, so \(A\) is strictly increasing. If the integral is negative for every such pair, then \(A\) is strictly decreasing.
A convenient sufficient condition is that the rate is continuous and positive throughout the interval: every interval of positive length then has positive signed area. A continuous rate that is nonnegative and is not identically zero on any subinterval also gives positive area on every interval of positive length, and hence strict increase. The corresponding conditions with nonpositive rates and negative area give strict decrease. These are sufficient ways to check strict monotonicity; the essential condition is the sign of the accumulated change on every interval.
The earlier material on monotonicity from the sign of the integrand relates the rate’s sign to the direction of accumulation. Here the focus is the consequence for a target: once strict monotonicity is established, no target value can occur at two inputs.
Worked Example: A Rate That Starts at Zero
Let \(r(t)=t\) on \(0\le t\le4\), and let \(A(0)=0\). The rate graph is a straight line from \((0,0)\) to \((4,4)\), so the accumulated amount at input \(x\) is the area of a triangle with base \(x\) and height \(x\):
Although the rate is zero at \(t=0\), it is positive for every \(t>0\). For any \(0\le x<y\le4\), the rate graph has positive area between \(x\) and \(y\), so \(A(y)-A(x)>0\). Thus \(A\) is strictly increasing on the whole domain.
To find whether the target \(2\) is reached, solve \(x^2/2=2\). Multiplying both sides by \(2\) gives \(x^2=4\). The solutions to that equation are \(x=2\) and \(x=-2\), but only \(x=2\) belongs to the domain \([0,4]\). Check the target using the graph: at \(x=2\), the triangle has base \(2\) and height \(2\), so its area is \((1/2)(2)(2)=2\). Therefore \(A(2)=2\), and strict increase proves this is the only input in the domain that reaches the target.
Using Piecewise Rate Graphs
A rate graph may change height at a breakpoint and still produce a strictly monotone accumulation function. To decide whether uniqueness applies, check the signed area between every pair of inputs. It is not necessary for the rate to be constant; it is the accumulated change over each interval that matters.
Worked Example: A Positive Piecewise-Constant Rate
Suppose the rate is \(2\) for \(0\le t\le2\), then \(1\) for \(2<t\le5\), and \(A(0)=0\). Both rate levels are positive. For \(0\le x\le2\), the graph gives a rectangle of width \(x\) and height \(2\), so \(A(x)=2x\). In particular, \(A(2)=2\cdot2=4\).
For \(2\le x\le5\), accumulate the first rectangle and then the rectangle from \(2\) to \(x\):
Consider the target \(5\). On the second part of the domain, \(x+2=5\), so \(x=3\). This input lies in \([2,5]\). Check its accumulated area: the first rectangle contributes \(2\cdot2=4\), and the next rectangle has width \(3-2=1\) and height \(1\), contributing \(1\cdot1=1\). The total is \(4+1=5\), as required. The rate is positive over every interval of positive length, so \(A\) is strictly increasing on \([0,5]\). Consequently, \(x=3\) is the only input in the domain that reaches \(5\).
A rate may instead be negative throughout. Then each additional interval contributes negative signed area, so accumulation strictly decreases. Because the direction is consistent, a target can still be reached at most once.
Worked Example: A Negative Piecewise-Constant Rate
Let \(r(t)=-1\) for \(0\le t\le1\), let \(r(t)=-2\) for \(1<t\le3\), and set \(A(0)=0\). On the first segment, the area is negative, so \(A(1)=-1\cdot1=-1\). For \(1\le x\le3\), the additional signed area is \(-2(x-1)\). Thus
For the target \(-3\), use the second part of the domain: \(1-2x=-3\), so \(-2x=-4\), and \(x=2\). Check directly from the graph: the first segment contributes \(-1\), and the segment from \(1\) to \(2\) contributes \((-2)(2-1)=-2\). Their sum is \(-1+(-2)=-3\). Every interval of positive length has negative signed area, so the accumulation is strictly decreasing. The target is therefore reached only at \(x=2\).
A Common Mistake: Confusing Nondecreasing With Strictly Increasing
A nonnegative rate does not always make accumulation strictly increasing. It guarantees that the signed area over an interval is nonnegative, so accumulation does not decrease. But if the rate is zero across a whole interval, the accumulated change there is zero and the accumulation has a flat section. A target on that flat section can have more than one input.
Worked Example: A Flat Section Gives Repeated Target Inputs
Let the rate be \(2\) on \(0\le t\le1\), \(0\) on \(1<t\le3\), and \(1\) on \(3<t\le4\), with \(A(0)=0\). The first segment contributes area \(2\cdot1=2\), so \(A(1)=2\). On the interval \([1,3]\), the rate is zero, and the additional area is \(0\cdot(3-1)=0\). Therefore \(A(x)=2\) for every \(x\in[1,3]\).
A mistaken argument would say, “The rate is nonnegative, so accumulation is strictly increasing, and the target \(2\) has only one input.” The conclusion is false: \(A(1)=2\), \(A(2)=2\), and \(A(3)=2\). The correction is that a nonnegative rate gives nondecreasing accumulation, not necessarily strictly increasing accumulation. Since this example has a flat interval, the strict-monotonicity uniqueness theorem does not apply.
How to Use the Uniqueness Result
When a problem asks whether a target accumulation can occur at more than one input, the main task is to establish strict monotonicity on the relevant domain. A rate graph offers a practical way to do this: compare the signed areas between arbitrary ordered inputs. If every such area is positive, accumulation is strictly increasing; if every such area is negative, it is strictly decreasing. Once that is shown, one verified target input is the only one.
The order of reasoning matters. First establish strict monotonicity on the full interval that contains the candidate inputs. Then verify that a proposed input actually gives the target. The theorem supplies uniqueness, not the numerical value of the input and not proof that the target is attained.
Check Your Understanding
Use the definitions and signed-area reasoning above to answer these questions.
- Why does strict increase rule out two different inputs \(x_1<x_2\) with \(A(x_1)=A(x_2)\)?
- What does the uniqueness theorem guarantee, and what does it not guarantee, about a target value?
- A continuous rate is nonnegative and is not identically zero on any subinterval. What does this imply about the accumulation function, and why?
- Can a nonnegative rate produce a target reached at several inputs? Describe the rate behavior that makes this possible.
- If a strictly decreasing accumulation has one input at which it equals a target, how many other inputs in its domain can equal that target?
