From Derivative Signs to Local Comparisons
A derivative sign describes how a function changes over an interval. A positive derivative means the function is increasing there; a negative derivative means it is decreasing. The First Derivative Test uses these facts to classify a point by looking at the signs immediately to its left and right.
The key question is why this interval information tells us how nearby function values compare with the value at the point itself. The connection is not merely a visual rule: it follows from the Mean Value Theorem, together with continuity at the point being classified.
In this tutorial, the sign conditions will hold throughout small intervals immediately to the left and right of \(c\). That is stronger than checking the derivative at one point: a single derivative value does not describe how the function behaves across a whole neighboring interval.
The Sign Patterns and Their Meaning
Suppose \(f\) is continuous at \(c\) and differentiable on both neighboring intervals \((c-\delta,c)\) and \((c,c+\delta)\), for some \(\delta>0\). The four possible strict sign patterns have a simple interpretation:
| Derivative on the left | Derivative on the right | Local behavior at \(c\) |
|---|---|---|
| Positive | Negative | Strict local maximum |
| Negative | Positive | Strict local minimum |
| Positive | Positive | Increasing on both sides; this sign pattern alone gives no extremum conclusion |
| Negative | Negative | Decreasing on both sides; this sign pattern alone gives no extremum conclusion |
The first two rows are the sign changes used by the First Derivative Test. A positive-to-negative change means the function rises toward \(c\) from the left and falls away from \(c\) on the right. A negative-to-positive change means it falls toward \(c\), then rises away from it.
Notice what the hypotheses do and do not require. The derivative signs need to hold on the neighboring open intervals; \(f'(c)\) itself need not exist. But the function must be continuous at \(c\), so the comparisons on either side connect to its actual value there.
Why the Theorem Follows
The Mean Value Theorem supplies the comparison on each side. In the form needed here, it says that if \(f\) is continuous on a closed interval \([u,v]\) and differentiable on its interior \((u,v)\), then for some \(w\in(u,v)\), \(f(v)-f(u)=f'(w)(v-u)\). Since \(v-u>0\), a positive derivative at \(w\) makes \(f(v)>f(u)\), while a negative derivative makes \(f(v)<f(u)\).
Apply the Mean Value Theorem to points strictly to the left of \(c\), or strictly to its right. The derivative sign determines which function value is larger.
Use continuity at \(c\) to relate the nearby values to \(f(c)\). Strict monotonicity ensures a strict comparison for every point other than \(c\).
Here is a complete proof for a positive-to-negative sign change. Choose any \(x\) with \(c-\delta<x<c\). Pick \(y\) with \(x<y<c\). On the interval from \(x\) to \(y\), the Mean Value Theorem applies, and \(f'\) is positive throughout its interior. Therefore \(f(y)>f(x)\).
For any \(z\) with \(y<z<c\), the same theorem gives \(f(z)>f(y)\), because the derivative remains positive. As \(z\) approaches \(c\) from the left, continuity at \(c\) gives \(f(z)\to f(c)\). Since every such \(f(z)\) exceeds \(f(y)\), it follows that \(f(c)\geq f(y)\). Consequently \(f(c)\geq f(y)>f(x)\), so \(f(x)<f(c)\).
Now take any \(x\) with \(c<x<c+\delta\). Choose \(y\) with \(c<y<x\). Since \(f'\) is negative to the right of \(c\), the Mean Value Theorem gives \(f(x)<f(y)\). Let points strictly between \(c\) and \(y\) approach \(c\) from the right. Their function values are greater than \(f(y)\), and continuity implies \(f(c)\geq f(y)\). Thus \(f(x)<f(y)\leq f(c)\). We have shown that every sufficiently nearby \(x\neq c\), on either side, satisfies \(f(x)<f(c)\). This is precisely a strict local maximum.
For a negative-to-positive sign change, the comparisons reverse. To the left, \(f\) decreases as \(x\) moves toward \(c\), so nearby left-hand values exceed \(f(c)\). To the right, \(f\) increases as \(x\) moves away from \(c\), so nearby right-hand values also exceed \(f(c)\). More explicitly, applying the same point comparisons with \(f'<0\) on the left and \(f'>0\) on the right gives \(f(x)>f(c)\) for every sufficiently close \(x\neq c\). Therefore \(c\) is a strict local minimum.
Worked Example: A Smooth Local Minimum
Worked Example: A Smooth Local Minimum
Classify the local behavior of \(f(x)=x^2-4x+5\) at \(c=2\) using derivative signs, and verify the result.
Differentiate:
If \(x<2\), then \(x-2<0\), so \(f'(x)<0\). If \(x>2\), then \(x-2>0\), so \(f'(x)>0\). The sign changes from negative to positive, so the theorem gives a strict local minimum at \(x=2\).
Evaluate the function there:
Check directly by completing the square:
For every \(x\neq2\), \((x-2)^2>0\), so \(f(x)>1=f(2)\). This confirms the strict local minimum (and, in this example, the same inequality holds for all \(x\)). Differentiating the completed-square expression also checks the derivative: \(\frac{d}{dx}\big((x-2)^2+1\big)=2(x-2)\), as before.
Worked Example: A Smooth Local Maximum
Worked Example: A Smooth Local Maximum
Classify \(g(x)=4-(x+1)^2\) at \(c=-1\), and verify the function value and sign conclusion.
Differentiate:
When \(x<-1\), \(x+1<0\), so \(-2(x+1)>0\). When \(x>-1\), \(x+1>0\), so \(-2(x+1)<0\). Thus the derivative changes from positive to negative, which proves that \(x=-1\) is a strict local maximum.
The value at the point is:
For any \(x\neq-1\), the square \((x+1)^2\) is positive. Hence \(g(x)=4-(x+1)^2<4=g(-1)\), directly verifying the strict maximum. As a numerical check, \(g(-2)=4-(-1)^2=3\) and \(g(0)=4-(1)^2=3\), both below \(4\). Differentiating the original expression gives \(-2(x+1)\), confirming the derivative used in the sign analysis.
Worked Example: A Local Minimum at a Nondifferentiable Point
Worked Example: A Local Minimum at a Nondifferentiable Point
Use neighboring derivative signs to classify \(h(x)=|x-2|+1\) at \(c=2\), even though the derivative at \(2\) does not exist.
For \(x<2\), \(|x-2|=2-x\), so \(h(x)=3-x\) and \(h'(x)=-1<0\). For \(x>2\), \(|x-2|=x-2\), so \(h(x)=x-1\) and \(h'(x)=1>0\). The sign changes from negative to positive. The function is continuous at \(2\), so the theorem applies and gives a strict local minimum.
At the point, \(h(2)=|0|+1=1\). If \(x<2\), then \(h(x)=3-x>1\); if \(x>2\), then \(h(x)=x-1>1\). These formulas verify that all nearby values other than \(h(2)\) are larger. The left-hand derivative is \(-1\) and the right-hand derivative is \(1\), so the derivative at \(2\) does not exist; that does not prevent the sign test from applying on the neighboring intervals.
Common Mistake: Reversing the Sign Pattern
A frequent error is to call a negative-to-positive change a maximum because the derivative changes sign. The direction of the change matters. For \(f(x)=x^2\), the derivative is \(f'(x)=2x\): it is negative for \(x<0\) and positive for \(x>0\). The function decreases toward \(0\), then increases away from \(0\), so the classification is a minimum, not a maximum.
The values confirm the correction: \(f(0)=0\), while \(f(-1)=(-1)^2=1\) and \(f(1)=1^2=1\). Both nearby values are greater than the value at \(0\). The correct correspondence is negative-to-positive for a minimum and positive-to-negative for a maximum.
What the Sign Test Does—and Does Not—Establish
The theorem gives a sufficient condition: when its hypotheses hold and the derivative has one of the two stated sign changes, the point is a strict local extremum. It is not a method for comparing all values of a function on its entire domain. The conclusion concerns only a neighborhood of \(c\); values far away may be larger or smaller.
Continuity at \(c\) is essential to the argument because it connects the behavior on the open intervals to the value \(f(c)\). Differentiability at \(c\), however, is not part of the sign-change hypothesis. The absolute-value example illustrates why this distinction matters: the function has the required continuity and neighboring derivative signs even though it has no derivative at the point.
In practice, the reasoning can be organized as follows:
- Choose a point \(c\) and intervals immediately to its left and right.
- Confirm continuity at \(c\), and determine the sign of \(f'\) throughout each interval.
- Translate each sign into increasing or decreasing behavior using the Mean Value Theorem.
- Compare the nearby function values with \(f(c)\) to classify the local extremum.
The sign chart is therefore more than a convenient visual aid. Its conclusion rests on a precise chain: derivative signs give interval comparisons; monotonic behavior on both sides, together with continuity at the point, gives comparisons with \(f(c)\); those comparisons are exactly the definition of a local maximum or minimum.
Check Your Understanding
Answer each question using the connection between derivative signs, monotonicity, and local comparisons.
- What sign change in \(f'\) gives a strict local maximum, and what sign change gives a strict local minimum?
- Why does a positive derivative on an interval imply that \(f\) is increasing there?
- In the sign-change theorem, why is continuity at \(c\) required even though differentiability at \(c\) is not?
- A function has \(f'(x)<0\) immediately to the left of \(c\) and \(f'(x)>0\) immediately to the right. What local classification follows, and why?
- What does the sign-change theorem establish about the neighborhood of \(c\), and what does it not establish about the whole domain?
