Why the Derivative’s Behavior on Both Sides Matters
A critical number is a candidate for a local maximum or minimum, but being a candidate does not guarantee that the function has an extremum there. The derivative’s value at one point is not enough: what matters is how the derivative behaves on either side of that point.
A derivative that changes from positive to negative indicates that the function switches from increasing to decreasing. A change from negative to positive indicates a switch from decreasing to increasing. If the derivative keeps the same strict sign on both sides, the function keeps moving in the same direction, so the critical number is neither a local maximum nor a local minimum.
Critical Numbers and Local Extrema
A number \(c\) is a critical number of a function \(f\) if \(c\) is in the domain of \(f\) and either \(f'(c)=0\) or \(f'(c)\) does not exist. In this tutorial, we focus on critical numbers where \(f\) is differentiable and \(f'(c)=0\); the derivative sign test described below applies under the stated conditions.
A function \(f\) has a local maximum at \(c\) if there is an interval around \(c\) in which \(f(c)\geq f(x)\) for every \(x\) in the domain. It has a local minimum at \(c\) if, in some interval around \(c\), \(f(c)\leq f(x)\) for every \(x\) in the domain. The word “local” means that the comparison is with nearby inputs, not necessarily with every input in the full domain.
When the inequalities are strict for every nearby \(x\ne c\), the maximum or minimum is called strict. A strict sign change in the derivative gives a strict local extremum.
The First-Derivative Sign Test
Suppose \(f\) is continuous on an interval around \(c\), differentiable on each side of \(c\), and its derivative has a constant strict sign on each side. Then the sign pattern gives the classification below.
| Sign of \(f'\) to the left | Sign of \(f'\) to the right | Conclusion at \(c\) |
|---|---|---|
| Positive | Negative | Strict local maximum |
| Negative | Positive | Strict local minimum |
| Positive | Positive | Neither; the function is increasing through \(c\) |
| Negative | Negative | Neither; the function is decreasing through \(c\) |
This test is a sufficient way to classify a critical number when its conditions hold. A zero derivative at \(c\) alone does not establish any of these outcomes. The sign chart is useful because it records the derivative’s behavior on intervals, rather than focusing only on the value at the critical number.
Why the Sign Test Works
The justification follows from the Mean Value Theorem. Recall that if a function is continuous on a closed interval and differentiable on its interior, then for some point in the interior its derivative equals the average rate of change across the interval.
Assume \(f\) is continuous on \([c-\delta,c+\delta]\) for some \(\delta>0\), differentiable on either side of \(c\), and \(f'(x)>0\) for \(c-\delta<x<c\), while \(f'(x)<0\) for \(c<x<c+\delta\). We show that \(c\) is a strict local maximum.
Choose any \(x\) with \(c-\delta<x<c\). The Mean Value Theorem applies to \([x,c]\), so there is a \(\xi\) with \(x<\xi<c\) such that \(f(c)-f(x)=f'(\xi)(c-x)\). Here \(f'(\xi)>0\) and \(c-x>0\), so \(f(c)-f(x)>0\). Thus \(f(x)<f(c)\).
Choose any \(x\) with \(c<x<c+\delta\). The Mean Value Theorem applies to \([c,x]\), so there is a \(\xi\) with \(c<\xi<x\) such that \(f(x)-f(c)=f'(\xi)(x-c)\). Here \(f'(\xi)<0\) and \(x-c>0\), so \(f(x)-f(c)<0\). Thus \(f(x)<f(c)\).
On both sides, every sufficiently nearby \(x\ne c\) satisfies \(f(x)<f(c)\). Therefore, \(c\) is a strict local maximum.
The same reasoning proves the other cases. If \(f'<0\) on the left and \(f'>0\) on the right, the comparisons give \(f(x)>f(c)\) on both sides, so \(c\) is a strict local minimum. If \(f'\) is positive on both sides, then points to the left have smaller values and points to the right have larger values. If \(f'\) is negative on both sides, points to the left have larger values and points to the right have smaller values. Either unchanged sign rules out both kinds of local extremum.
Worked Example: A Negative-to-Positive Change
Worked Example: A Negative-to-Positive Change
Classify the critical number of \(f(x)=(x-2)^2\).
First differentiate: \(f'(x)=2(x-2)\). Set the derivative equal to zero: \(2(x-2)=0\), so \(x-2=0\), and the critical number is \(c=2\).
For \(x<2\), \(x-2<0\), so \(f'(x)=2(x-2)<0\). For \(x>2\), \(x-2>0\), so \(f'(x)=2(x-2)>0\). The sign changes from negative to positive, which gives a strict local minimum.
The function value is \(f(2)=(2-2)^2=0\). For a direct check, \(f(1)=(1-2)^2=1\) and \(f(3)=(3-2)^2=1\), both greater than \(f(2)\). Also, \(f'(2)=2(2-2)=0\), confirming that \(2\) is a critical number.
Worked Example: A Positive-to-Negative Change
Worked Example: A Positive-to-Negative Change
Classify the critical number of \(g(x)=-(x+1)^2\).
Differentiate: \(g'(x)=-2(x+1)\). Solve \(g'(x)=0\): \(-2(x+1)=0\), so \(x+1=0\), and \(c=-1\).
For \(x<-1\), \(x+1<0\), so \(-2(x+1)>0\). For \(x>-1\), \(x+1>0\), so \(-2(x+1)<0\). The derivative changes from positive to negative, so \(g\) has a strict local maximum at \(x=-1\).
The function value is \(g(-1)=-((-1)+1)^2=0\). Check nearby values: \(g(-2)=-((-2)+1)^2=-1\) and \(g(0)=-(0+1)^2=-1\). Both are below \(g(-1)\). Substitution also gives \(g'(-1)=-2((-1)+1)=0\), as required for this critical number.
Worked Example: No Sign Change
Worked Example: No Sign Change
Determine whether \(h(x)=x^3\) has a local extremum at its critical number.
Differentiate: \(h'(x)=3x^2\). Solving \(3x^2=0\) gives \(x=0\), so \(c=0\) is the critical number.
For \(x<0\), \(x^2>0\), so \(3x^2>0\). For \(x>0\), \(x^2>0\), so \(3x^2>0\) as well. The derivative is positive on both sides. There is no sign change, so \(0\) is neither a local maximum nor a local minimum.
Indeed, \(h(-1)=(-1)^3=-1\), \(h(0)=0^3=0\), and \(h(1)=1^3=1\). Nearby values on the left are below \(h(0)\), while nearby values on the right are above it. Also, \(h'(0)=3(0)^2=0\), verifying the critical-number calculation.
A Common Mistake: Treating Every Critical Number as an Extremum
A frequent error is to see \(f'(c)=0\) and immediately label \(c\) a maximum or minimum. For \(h(x)=x^3\), an incorrect calculation might say, “\(h'(0)=0\), so \(x=0\) is a local minimum.” The conclusion does not follow from that equation: it identifies a critical number but does not classify it.
The correction is to check both sides. Since \(h'(x)=3x^2>0\) for every \(x\ne0\), the derivative is positive both to the left and to the right of zero. Thus the function increases through zero and has no local extremum there. The derivative’s value at the critical number identifies where to investigate; its signs around that number determine the classification.
Using the Test Carefully
To apply the test, first identify a critical number in the domain. Then determine the sign of the derivative on an interval immediately to its left and on an interval immediately to its right. A sign chart can organize this information; the important comparison is between the two side signs, not between either sign and the value \(f'(c)=0\).
The sign test gives a local conclusion. A local maximum need not be the greatest function value on the entire domain, and a local minimum need not be the least. Also, the test as stated relies on strict derivative signs throughout the neighboring intervals and continuity at \(c\). If those conditions are not established, the pattern alone does not justify the conclusion.
Check Your Understanding
For each question, focus on the derivative signs immediately to the left and right of the critical number.
- If \(f'\) changes from positive to negative at \(c\), what type of local behavior occurs at \(c\)?
- If \(f'\) changes from negative to positive at \(c\), what type of local behavior occurs at \(c\)?
- If \(f'\) is negative on both sides of a critical number, is that number a local maximum, a local minimum, or neither?
- Why does \(f'(c)=0\) by itself not prove that \(f\) has a local extremum at \(c\)?
- Which theorem justifies comparing \(f(x)\) with \(f(c)\) using the derivative’s sign on an interval?
