How a Rate Change Creates an Extreme
An accumulation function records the net amount accumulated from a fixed starting point. Its value rises where its integrand is positive and falls where its integrand is negative. If the integrand changes from positive to negative, the accumulation function switches from rising to falling, suggesting a local maximum. A change from negative to positive suggests a local minimum.
The key is to compare the accumulation at the candidate point with its values nearby. Signed areas make this comparison precise: the sign of the area between two inputs tells us whether the accumulation has increased or decreased between them.
For the discussion below, \(c\) is an interior point of the domain. We will use the interval-increment identity: for any \(x\) and \(y\) in the domain, \(A(y)-A(x)=\int_x^y f(t)\,dt\). This identity says that the change in accumulation is exactly the signed area over the interval traversed.
The Sign-Change Rule
Suppose the integrand is strictly positive just to the left of \(c\) and strictly negative just to the right. Moving toward \(c\) from the left adds positive signed area, so the accumulation increases. Moving right from \(c\) adds negative signed area, so the accumulation decreases. The point \(c\) is therefore a strict local maximum.
Reversing the signs reverses the conclusion. If the integrand is negative just to the left and positive just to the right, the accumulation decreases toward \(c\) and increases after \(c\); this gives a strict local minimum.
The result requires the stated sign pattern on whole intervals next to \(c\), not merely at a few sampled points. The integrand’s value at the single point \(c\) does not affect the signed areas, so the theorem does not require \(f(c)=0\). When \(f\) is continuous at \(c\), a strict sign change across \(c\) does imply \(f(c)=0\); the important information for classifying the accumulation extreme is the sign on each side.
Why the Rule Works
Here is a direct proof using signed areas. Choose any \(x<c\) sufficiently close to \(c\). By the interval-increment identity, \(A(c)-A(x)\) is the integral of \(f\) from \(x\) to \(c\). If \(f\) is positive throughout that interval, the integral is positive, so \(A(c)>A(x)\). For any \(x>c\) sufficiently close to \(c\), \(A(x)-A(c)\) is the integral from \(c\) to \(x\). If \(f\) is negative there, that integral is negative, so \(A(x)<A(c)\). Thus every nearby value other than \(A(c)\) is smaller: \(c\) is a strict local maximum. The minimum case follows by reversing both signs and inequalities.
Positive integrand means accumulation rises as the input approaches \(c\) from the left; negative integrand means it falls.
Positive integrand means accumulation rises as the input moves right from \(c\); negative integrand means it falls.
Rising then falling gives a local maximum; falling then rising gives a local minimum. Matching directions do not give an extreme by this sign-change test.
Worked Example: A Local Minimum From a Negative-to-Positive Change
Worked Example: A Local Minimum From a Negative-to-Positive Change
Let \(A(x)=\int_0^x (t-2)\,dt\). Locate and verify the local extreme suggested by the integrand.
The integrand \(f(t)=t-2\) is negative when \(t<2\), equals zero at \(t=2\), and is positive when \(t>2\). Its sign changes from negative to positive, so \(x=2\) is a candidate local minimum.
First find the accumulation value at the candidate. An antiderivative of \(t-2\) is \(\frac{t^2}{2}-2t\), so
To verify the minimum on the left, take \(0<h<2\). The interval-increment identity gives
On the right, for \(h>0\),
Both nearby values are greater than \(A(2)=-2\), so \(x=2\) is a strict local minimum. As a check, \(A(x)=\frac{x^2}{2}-2x\), and substituting \(x=2-h\) or \(x=2+h\) gives \(-2+\frac{h^2}{2}\), in agreement with the signed-area calculations.
Worked Example: A Local Maximum From a Positive-to-Negative Change
Worked Example: A Local Maximum From a Positive-to-Negative Change
Let \(B(x)=\int_0^x(3-t)\,dt\). Determine and verify the local extreme at the sign-change point.
The integrand \(3-t\) is positive for \(t<3\) and negative for \(t>3\). The positive-to-negative change makes \(x=3\) a candidate local maximum. Its accumulation value is
For \(0<h<3\), the change from the left-hand input \(3-h\) to \(3\) is
Therefore \(B(3-h)<B(3)\). On the right,
so \(B(3+h)<B(3)\) as well. This verifies a strict local maximum. Indeed, direct evaluation gives \(B(x)=3x-\frac{x^2}{2}\), and substituting \(x=3-h\) and \(x=3+h\) yields \(\frac{9}{2}-\frac{h^2}{2}\) in both cases, confirming the signed-area comparisons.
Worked Example: Classifying More Than One Sign Change
Worked Example: Classifying More Than One Sign Change
Let \(C(x)=\int_0^x(t-1)(t-3)\,dt\). Find the local extremes suggested by the integrand and verify their accumulation values.
The integrand is \(f(t)=(t-1)(t-3)\). For \(t<1\), both factors are negative, so their product is positive. For \(1<t<3\), the first factor is positive and the second negative, so the product is negative. For \(t>3\), both factors are positive, so the product is positive. Thus the sign changes positive-to-negative at \(1\), and negative-to-positive at \(3\): these are candidates for a local maximum and a local minimum, respectively.
Expand the integrand and integrate:
The values at the candidates are
For \(0<h<1\), the signed area from \(1\) to \(1+h\) is negative, because the integrand is negative between \(1\) and \(3\). More explicitly, writing \(t=1+u\) gives \(f(1+u)=u(u-2)\), so
For \(0<h<1\), the interval from \(1-h\) to \(1\) has positive integrand; direct substitution \(t=1-u\) gives
Both neighboring values are below \(C(1)\), verifying the strict local maximum. At \(3\), the signs on the two sides are negative and positive. For \(0<h<1\), substitution \(t=3+u\) gives \(f(3+u)=u(u+2)\), and therefore
On the left, substitution \(t=3-u\) gives \(f(3-u)=u(u-2)\), hence
Both nearby values are above \(C(3)=0\), verifying the strict local minimum. The calculations confirm the sign-change rule and show why each side must be checked against the candidate value.
A Zero Is Not Necessarily an Extreme
A common mistake is to assume that every input where \(f(c)=0\) produces a local extreme of the accumulation function. A zero is only a candidate: the signs on either side determine whether the accumulation changes direction.
Worked Example: A Zero Without a Sign Change
Consider \(D(x)=\int_0^x(t-1)^2\,dt\). Its integrand is zero at \(t=1\), but it is positive on both sides of \(1\). For \(h>0\), compare nearby accumulation values with \(D(1)\):
where \(u=t-1\). For the input on the left,
The accumulation is below \(D(1)\) on the left and above it on the right. Thus \(D\) is still increasing through \(1\), and \(1\) is neither a local maximum nor a local minimum. The integrand vanishes there but does not change sign.
Another important distinction is between the sign of the integrand and the sign of the accumulated value. An accumulation function can have a negative value while increasing, or a positive value while decreasing. The local direction is determined by signed areas between nearby inputs, not by whether the graph of the accumulation is above or below zero.
A Practical Check
To locate and verify local extremes, first identify points where the integrand changes sign. Then inspect one interval on each side of each candidate. Do not infer an extreme from a zero alone, and do not use the sign of the accumulation value in place of the integrand’s sign. Finally, verify the conclusion by comparing the signed areas from the candidate to a nearby input on each side.
Check Your Understanding
Use the integrand’s signs and signed-area comparisons to answer these questions.
- If an integrand is negative immediately to the left of \(c\) and positive immediately to the right, what type of local extreme does the accumulation function have at \(c\)?
- Why does a positive-to-negative sign change give values below \(A(c)\) on both sides of \(c\)?
- Does \(f(c)=0\) by itself establish a local extreme of \(A(x)=\int_a^x f(t)\,dt\)? Explain what else must be checked.
- For \(f(t)=(t-2)^2\), classify the accumulation at \(c=2\), or explain why the sign-change test gives no extreme there.
- Can an accumulation value be negative while the accumulation function is increasing? Explain using the sign of the integrand.
