From Endpoint Values to a Target Input
An accumulation function records the net effect of a rate from a chosen starting input up to a variable input. If the rate is bounded, the accumulation changes continuously as the input changes. That continuity has an important consequence: when the accumulation is below a target at one endpoint and above it at another, it must equal the target somewhere in between.
This is an existence result. It can guarantee that a target is reached even when we cannot identify the exact input where it is reached. The key is to check the accumulation values at the endpoints of an interval, not merely the values of the rate.
For a bounded rate, the accumulation function is continuous. In particular, if \(|r(t)|\le M\) throughout the interval, then for inputs \(x\) and \(y\) in that interval, $$ |A(x)-A(y)|=\left|\int_y^x r(t)\,dt\right|\le M|x-y|. $$ This bound means that changing the input by a small amount can change the accumulation by only a small amount. Continuity is the property that lets us pass from endpoint comparisons to a value attained between the endpoints.
The Intermediate Value Theorem for Accumulations
The Intermediate Value Theorem applies to any function continuous on a closed interval. An accumulation function with a bounded rate is continuous there, so the theorem applies to it.
The word strictly matters for the conclusion \(c\in(u,v)\). If the target equals an endpoint value, that endpoint itself already reaches the target, but the theorem does not then guarantee a solution in the open interval.
Why the Theorem Gives an Input
Here is a proof of the theorem in the case \(A(u)<T<A(v)\). It makes explicit why continuity and the endpoint inequalities are enough. Let $$ S=\{x\in[u,v]:A(x)\le T\}. $$ Because \(A(u)<T\), the set \(S\) is nonempty. It is bounded above by \(v\), so it has a supremum \(c=\sup S\).
Since \(A(u)<T\) and \(A\) is continuous, values of \(A\) remain below \(T\) for inputs sufficiently close to \(u\). Thus \(S\) contains inputs greater than \(u\), so \(c>u\). Since \(A(v)>T\), continuity makes \(A(x)>T\) for inputs sufficiently close to \(v\). Those inputs are not in \(S\), so \(c<v\).
If \(A(c)<T\), continuity would make \(A(x)<T\) for inputs \(x\) sufficiently close to \(c\), including some \(x>c\). Such an \(x\) would belong to \(S\), contradicting that \(c\) is an upper bound for \(S\).
If \(A(c)>T\), continuity would make \(A(x)>T\) for inputs sufficiently close to \(c\). But because \(c=\sup S\), there are points of \(S\) arbitrarily close to \(c\) from below. At each such point \(x\in S\), the definition of \(S\) gives \(A(x)\le T\), a contradiction.
The only remaining possibility is \(A(c)=T\). Since \(u<c<v\), the target is reached at an input inside the interval.
If the endpoint values are in the reverse order, apply the result just proved to the continuous function \(-A\) and the target \(-T\). This covers both possible ways for the endpoint values to straddle the target.
Worked Example: Endpoint Values Straddle the Target in Increasing Order
Worked Example: Endpoint Values Straddle the Target in Increasing Order
Let the rate be \(r(t)=1\) on \([0,2)\) and \(r(t)=-\frac14\) on \([2,4]\). Define \(A(x)=\int_0^x r(t)\,dt\). Does the accumulation reach the target \(T=\frac54\) somewhere between \(0\) and \(4\)?
First, \(A(0)=0\), since the interval of accumulation has zero length. At \(x=4\), the first rate contributes a rectangle of signed area \(1\cdot2=2\). The second contributes a rectangle of signed area \(-\frac14\cdot2=-\frac12\). Therefore $$ A(4)=2-\frac12=\frac32. $$ The endpoint comparison is $$ A(0)=0<\frac54<\frac32=A(4). $$ The rate is bounded, so \(A\) is continuous on \([0,4]\). The Intermediate Value Theorem therefore guarantees some \(c\in(0,4)\) such that \(A(c)=\frac54\).
Check: The computed endpoint values differ by the net signed area \(2-\frac12=\frac32\), and the target satisfies \(0<\frac54<\frac32\). These are exactly the continuity and bracketing conditions required.
The theorem guarantees an input, but the endpoint comparison alone does not calculate it. That distinction is useful: an existence proof can be complete even when it does not give a numerical or exact formula for the input.
Worked Example: Endpoint Values Straddle the Target in Reverse Order
Worked Example: Endpoint Values Straddle the Target in Reverse Order
Let \(r(t)=-1\) on \([0,2)\) and \(r(t)=\frac14\) on \([2,4]\), with \(A(x)=\int_0^x r(t)\,dt\). Show that the accumulation reaches \(T=-1\) on \([0,4]\).
At the starting input, \(A(0)=0\). From \(0\) to \(2\), the signed area is \((-1)\cdot2=-2\). From \(2\) to \(4\), the signed area is \(\frac14\cdot2=\frac12\). Thus $$ A(4)=-2+\frac12=-\frac32. $$ The target lies between the endpoint values in reverse order: $$ A(4)=-\frac32<-1<0=A(0). $$ The rate is bounded, so \(A\) is continuous. The Intermediate Value Theorem guarantees an input \(c\in(0,4)\) with \(A(c)=-1\).
Check: The positive rate on \([2,4]\) contributes \(+\frac12\), not a negative area. With that sign, the endpoint value is \(-\frac32\), and \(-\frac32<-1<0\) verifies the required bracketing.
Using an Interval That Does Not Start at the Base Input
The interval tested by the theorem need not begin at the base input of the accumulation function. Once the accumulation values at the two selected endpoints are known, it is their comparison with the target that matters.
Worked Example: Bracketing a Target Away from the Base Input
Define \(A(x)=\int_{-1}^{x}r(t)\,dt\), where \(r(t)=2\) on \([-1,0]\), \(r(t)=2-\frac32t\) on \([0,2]\), and \(r(t)=-1\) on \([2,3]\). Show that \(A\) reaches \(T=\frac52\) somewhere between \(0\) and \(2\).
From \(-1\) to \(0\), the rate is constant at \(2\) over a length of \(1\). Hence $$ A(0)=2\cdot1=2. $$ On \([0,2]\), the rate is linear, with endpoint values \(r(0)=2\) and \(r(2)=2-3=-1\). Its signed area is the trapezoid area: the average endpoint height, \(\frac{2+(-1)}{2}=\frac12\), multiplied by the width \(2\). Therefore the additional accumulation is \(\frac12\cdot2=1\), and $$ A(2)=A(0)+1=2+1=3. $$ Now compare the target with these values: $$ A(0)=2<\frac52<3=A(2). $$ The rate is bounded on \([0,2]\), so \(A\) is continuous there. The theorem guarantees some \(c\in(0,2)\) for which \(A(c)=\frac52\).
Check: The accumulated value at \(2\) includes both the earlier area \(2\) from \([-1,0]\) and the area \(1\) from \([0,2]\), giving \(3\). The target is strictly between the two values on the interval being tested.
A Common Sign Error
Signed area determines the endpoint values. A positive rate contributes positive accumulation, even if it occurs after a negative rate. Losing that sign can create a false bracketing claim.
For the rate in the reverse-order example, a mistaken calculation might treat the positive contribution on \([2,4]\) as negative: $$ A(4)\ \text{(incorrect)}=-2-\frac12=-\frac52. $$ That incorrect endpoint would appear to put \(T=-2\) between \(A(4)=-\frac52\) and \(A(0)=0\). But the positive rate contributes \(+\frac12\), so the correct calculation is $$ A(4)=-2+\frac12=-\frac32. $$ Now \(-2\) is not between \(-\frac32\) and \(0\). The endpoint test therefore gives no Intermediate Value Theorem guarantee for that target on \([0,4]\). The correction is to preserve the sign of every area before comparing with the target.
What the Endpoint Test Does—and Does Not—Show
To use the theorem for an accumulation function on \([u,v]\), check these facts:
- The rate is bounded on the interval, ensuring that the accumulation function is continuous there.
- The target lies strictly between the two endpoint accumulation values.
- The interval and accumulation values refer to the same function and the same inputs.
When these conditions hold, at least one input in \((u,v)\) reaches the target. The theorem does not say how many such inputs there are, nor does it identify their exact locations. Conversely, if the endpoint values are on the same side of the target, this particular endpoint test does not establish that the target is reached. The accumulation might still reach it between the endpoints; the endpoint comparison alone simply does not decide the question.
Check Your Understanding
Use endpoint values and continuity to decide what the Intermediate Value Theorem guarantees in each situation.
- If \(A\) is continuous on \([1,5]\), \(A(1)=3\), and \(A(5)=8\), does \(A\) reach the target \(6\) at an input strictly between \(1\) and \(5\)? State the reason.
- If \(A(2)=4\) and \(A(7)=-1\), which endpoint order should be used to test whether \(A\) reaches the target \(2\)?
- A bounded rate defines an accumulation \(A(x)=\int_0^x r(t)\,dt\). What fact about \(A\) allows the Intermediate Value Theorem to be applied on a closed interval?
- If the target equals \(A(u)\), what input is already known to reach it? Does the strict-between-endpoints theorem guarantee an input in \((u,v)\) in this case?
- If both endpoint accumulation values are below a target, what can you conclude from the endpoint test alone?
