Why a Sign Chart Helps
A polynomial can have several critical numbers, and not every critical number is a local maximum or minimum. A derivative sign chart organizes the information: it shows where the function increases or decreases and whether that behavior changes at each critical number.
For a polynomial, the derivative exists everywhere. Thus its critical numbers are the real numbers where its derivative is zero. Once those numbers are found, the intervals between them can be checked one at a time. The sign of the derivative on each interval determines the direction of change of the function.
Definitions and the Sign-Change Rule
A function \(f\) has a local maximum at \(x=c\) if there is an open interval around \(c\) on which \(f(x)\leq f(c)\). It has a local minimum at \(c\) if there is an open interval around \(c\) on which \(f(x)\geq f(c)\). These definitions are non-strict, so equality at other nearby points is allowed.
A critical number of a function is a number in its domain where its derivative is zero or does not exist. Since every polynomial is differentiable on the real line, a polynomial’s critical numbers are precisely the real solutions of \(f'(x)=0\).
Here, “immediately” means throughout some interval just to that side of \(c\), with no other critical number in that interval. A nonconstant polynomial has only finitely many critical numbers, so such intervals can be chosen around each one. The theorem relies on definite signs on both sides; it does not apply to the constant-polynomial exception discussed below.
Why the Sign-Change Rule Works
The justification uses the Mean Value Theorem. A polynomial is continuous on every closed interval and differentiable on every open interval, so the theorem applies between any two real numbers. If \(f'\) is positive throughout an interval, then for any \(u<v\) in that interval, the Mean Value Theorem gives a number \(r\) between them such that
Because \(v-u>0\), this implies \(f(v)>f(u)\): \(f\) is strictly increasing there. If \(f'\) is negative, the same equation gives \(f(v)<f(u)\), so \(f\) is strictly decreasing.
Suppose now that \(f'\) is positive just left of \(c\) and negative just right of \(c\). On the left, \(f\) increases toward \(c\); on the right, it decreases away from \(c\). More explicitly, for a nearby \(x<c\), the Mean Value Theorem on \([x,c]\) gives \(f(c)-f(x)>0\), and for a nearby \(y>c\), it gives \(f(y)-f(c)<0\). Thus \(f(x)<f(c)\) and \(f(y)<f(c)\): \(c\) is a local maximum. Reversing both signs gives \(f(x)>f(c)\) and \(f(y)>f(c)\), establishing a local minimum. If the signs are positive on both sides, the function increases through \(c\); if negative on both sides, it decreases through \(c\). Neither pattern creates a local extremum.
Building and Reading the Chart
For a factored derivative, the real zeros divide the number line into intervals. On each interval, \(f'\) is continuous and does not equal zero, so its sign is constant. You can determine that sign by testing one convenient value in the interval, or by tracking the signs of the factors.
Factor \(f'(x)\), solve \(f'(x)=0\), and list the distinct real solutions in increasing order.
Use the critical numbers to partition the real line into open intervals.
Substitute a test value from each interval into the factored derivative, or determine the signs of its factors.
Read the signs on its left and right; then evaluate \(f(c)\) to give the point \((c,f(c))\).
A critical number is only a candidate for an extremum. The sign change—not merely the fact that \(f'(c)=0\)—determines the classification. Also, sign charts classify local behavior; they do not by themselves establish whether an extremum is an absolute maximum or minimum on an entire domain.
Worked Example: Two Critical Numbers
Worked Example: Two Critical Numbers
Classify the critical numbers of \(f(x)=x^3-3x\), and give the corresponding local-extremum points.
Differentiate and factor:
Set the derivative equal to zero:
These values divide the number line into \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). Test \(x=-2,0,2\), respectively:
| Interval or number | Derivative sign or conclusion |
|---|---|
| \((-\infty,-1)\) | \(f'(x)>0\) |
| \(x=-1\) | Positive to negative: local maximum |
| \((-1,1)\) | \(f'(x)<0\) |
| \(x=1\) | Negative to positive: local minimum |
| \((1,\infty)\) | \(f'(x)>0\) |
Evaluate the original function at both critical numbers:
Therefore, the local maximum point is \((-1,2)\), and the local minimum point is \((1,-2)\).
Check: Differentiating the original function gives \(3x^2-3\), which factors as \(3(x-1)(x+1)\), as used in the chart. The signs show increase, then decrease, then increase, matching the maximum and minimum classifications.
Worked Example: Several Extrema in One Polynomial
Worked Example: Three Critical Numbers
Classify the critical numbers of \(g(x)=x^4-2x^2\), and find the local-extremum points.
Differentiate and factor:
The critical numbers are \(x=-1,0,1\). Test one value in each interval:
Thus the signs across the intervals are negative, positive, negative, positive. At \(x=-1\), the sign changes from negative to positive, so there is a local minimum. At \(x=0\), it changes from positive to negative, so there is a local maximum. At \(x=1\), it changes from negative to positive, so there is a local minimum.
Evaluate \(g\) at each critical number:
The local minima are \((-1,-1)\) and \((1,-1)\); the local maximum is \((0,0)\).
Check: Differentiating \(x^4-2x^2\) gives \(4x^3-4x\), and expanding \(4x(x-1)(x+1)\) gives \(4x(x^2-1)=4x^3-4x\). The derivative signs on the four intervals agree with the stated changes in behavior.
Worked Example: A Factored Derivative with Three Distinct Roots
Worked Example: Classifying Three Critical Numbers
Let
Find and classify its critical numbers, then report the local-extremum points.
Differentiate each term:
The roots are \(-2,1,3\), so test values \(-3,0,2,4\) in the four intervals:
The sign pattern is negative, positive, negative, positive. Hence \(-2\) is a local minimum, \(1\) is a local maximum, and \(3\) is a local minimum.
Now calculate the function values, keeping the arithmetic exact:
Therefore the local minimum points are \(\left(-2,-\frac{38}{3}\right)\) and \(\left(3,-\frac94\right)\), and the local maximum point is \(\left(1,\frac{37}{12}\right)\).
Check: Differentiating \(h\) gives \(x^3-2x^2-5x+6\). Expanding the factored derivative gives \((x+2)(x-1)(x-3)=(x+2)(x^2-4x+3)=x^3-2x^2-5x+6\). The factorization and sign chart are consistent.
A Common Mistake: Treating Every Critical Number as an Extremum
Finding \(f'(c)=0\) does not by itself prove that \(c\) is a local maximum or minimum. The signs on both sides must be checked. For example, in the previous example \(x=1\) is a critical number because \(h'(1)=0\), but its classification depends on the positive-to-negative sign change, not on the equation \(h'(1)=0\) alone.
There is also an important exception to broad claims about “no sign change.” If \(f\) is a nonconstant polynomial and its derivative has the same definite nonzero sign on both sides of an isolated critical number, that number is not a local extremum. But the constant polynomial \(f(x)=0\) has derivative zero everywhere, and every real number is both a local maximum and a local minimum under the non-strict definitions. Its derivative does not have a definite nonzero sign on either side, so the sign-change theorem above does not apply.
Check Your Understanding
Use factored derivatives and sign changes to answer these questions.
- If \(f'(x)=(x+1)(x-2)\), what are the critical numbers, and what is the derivative sign on each interval they determine?
- A polynomial’s derivative changes from negative to positive at \(x=c\). What type of local extremum occurs there?
- For \(p(x)=x^3-3x\), what are the local-extremum points?
- Why does \(f'(c)=0\) alone not establish that a polynomial has a local extremum at \(c\)?
- What is the local-extremum status of every real number for a constant polynomial, using the non-strict definitions?
