Why Multiplicity Predicts a Sign Change
A polynomial derivative can vanish at a point without changing sign there. When it does change sign, the direction of the change determines whether the original function has a local maximum or minimum. For a factored derivative, the multiplicity of the zero gives a quick way to predict whether a sign change occurs.
The key distinction is parity: a zero of odd multiplicity reverses the derivative’s sign as the input passes through it, while a zero of even multiplicity does not. The other factors in the derivative determine which sign occurs on each side, and therefore which type of extremum is present when a sign change does occur.
Multiplicity of a Zero
Suppose \(P\) is a polynomial and \(r\) is a zero of \(P'\). The multiplicity of this zero is the largest positive integer \(m\) for which \((x-r)^m\) is a factor of \(P'(x)\). Equivalently, we can write \(P'(x)=(x-r)^m h(x)\), where \(h\) is a polynomial and \(h(r)\ne 0\).
The condition \(h(r)\ne0\) matters: it means that all copies of the factor \((x-r)\) have been accounted for. Since \(h\) is continuous and nonzero at \(r\), it keeps the same sign on some interval around \(r\). Thus, sufficiently close to \(r\), the sign behavior of \(P'(x)\) is controlled by \((x-r)^m\), together with the fixed sign of \(h(x)\).
Multiplicity here refers to a zero of the derivative. It need not be the multiplicity of \(r\) as a zero of the original polynomial \(P\). Those are different questions: a point can be a zero of \(P'\) even when \(P(r)\ne0\).
The Sign-Change Rule
Why the Rule Works
Write \(P'(x)=(x-r)^m h(x)\), where \(h(r)\ne0\). Continuity gives a small interval around \(r\) on which \(h(x)\) never vanishes and has the same sign on both sides. So it remains to compare \((x-r)^m\) immediately to the left and right of \(r\).
If \(m\) is even, raising either a negative or a positive number to the power \(m\) gives a positive result. Hence \((x-r)^m\) has the same sign on both sides. Multiplication by \(h(x)\), whose sign is fixed nearby, preserves that agreement.
If \(m\) is odd, \((x-r)^m\) is negative to the left and positive to the right. Multiplication by a factor that keeps the same nonzero sign cannot remove this reversal. Therefore \(P'\) changes sign. This proves both directions of the rule.
The rule tells us whether a sign change occurs, but not its direction. To find that, determine the sign of the remaining factor \(h(x)\) near \(r\). If it is positive, an odd-multiplicity zero changes the derivative from negative to positive. If it is negative, the change is from positive to negative.
From Sign Changes to Local Extrema
For a polynomial, the derivative is continuous. Where \(P'\) is positive on an interval, \(P\) is increasing there; where \(P'\) is negative, \(P\) is decreasing there. The mean value theorem justifies this connection: for any two points \(a<b\) in such an interval, there is a point \(c\) between them with \(P(b)-P(a)=P'(c)(b-a)\). Since \(b-a>0\), the sign of the difference is the sign of \(P'(c)\).
Consequently, if \(P'\) changes from negative to positive at \(r\), \(P\) decreases toward \(r\) and increases after it. This gives a local minimum. A change from positive to negative gives a local maximum. If the derivative has the same nonzero sign immediately on both sides, the function is increasing on both sides or decreasing on both sides, so there is no local extremum at \(r\).
These conclusions concern an interior point of the domain. Polynomials are defined and differentiable on all of \(\mathbb{R}\), so every real number considered here is an interior point. The point \(r\) is stationary because \(P'(r)=0\); being stationary alone does not guarantee a local extremum.
Worked Examples
Worked Example: An Even-Multiplicity Zero Without an Extremum
Let \(P(x)=x^3-3x^2+3x\). Find its stationary point and decide whether it is a local extremum.
First differentiate:
The derivative is zero when \(3(x-1)^2=0\), so \(x=1\). The zero has multiplicity \(2\), which is even. The factor \(3\) is positive, and \((x-1)^2\) is positive whenever \(x\ne1\). Thus \(P'(x)>0\) immediately to the left and to the right of \(1\). There is no sign change, so \(P\) is increasing on both sides and has no local extremum at \(1\).
Check the derivative by differentiating the original polynomial: \(\frac{d}{dx}(x^3-3x^2+3x)=3x^2-6x+3=3(x-1)^2\), as required. Also, \(P(1)=1-3+3=1\); the stationary point is not a zero of \(P\).
Worked Example: An Odd-Multiplicity Zero That Gives a Minimum
Let \(Q(x)=\frac{1}{2}(x+1)^4\). Classify its stationary point using the multiplicity of the derivative’s zero.
Differentiate using the chain rule:
The derivative vanishes at \(x=-1\), where its zero has multiplicity \(3\), which is odd. For \(x<-1\), \(x+1<0\), so \(Q'(x)<0\). For \(x>-1\), \(x+1>0\), so \(Q'(x)>0\). The sign changes from negative to positive; therefore \(Q\) has a local minimum at \(x=-1\).
To check, \(Q(-1)=\frac12(0)^4=0\), and \(Q(x)=\frac12(x+1)^4\ge0\) for every real \(x\). This confirms the minimum. Differentiating the proposed derivative check gives \(\frac{d}{dx}\bigl(\frac12(x+1)^4\bigr)=2(x+1)^3\), matching the expression used for the sign test.
Worked Example: Classifying Multiple Zeros in One Derivative
Suppose \(R(x)=\frac{x^4}{4}-\frac{3x^2}{2}-2x\). Find the stationary points and classify each one.
Differentiate:
Factor the derivative by checking the factors \(x+1\) and \(x-2\):
Thus \(R'(x)=(x+1)^2(x-2)\), and its stationary points are \(x=-1\) and \(x=2\). At \(-1\), the zero has multiplicity \(2\). Near \(-1\), the squared factor is positive except at the zero, while \(x-2<0\). Therefore \(R'\) is negative on both sides of \(-1\). There is no sign change and no local extremum there.
At \(2\), the factor \(x-2\) has multiplicity \(1\), which is odd. For \(x<2\) close to \(2\), \((x+1)^2>0\) and \(x-2<0\), so \(R'(x)<0\). For \(x>2\), both factors are positive, so \(R'(x)>0\). The change from negative to positive means that \(R\) has a local minimum at \(2\).
Check by differentiation: \(\frac{d}{dx}\bigl(\frac{x^4}{4}\bigr)=x^3\), \(\frac{d}{dx}\bigl(-\frac{3x^2}{2}\bigr)=-3x\), and \(\frac{d}{dx}(-2x)=-2\). Therefore \(R'(x)=x^3-3x-2=(x+1)^2(x-2)\), confirming both the factorization and the points analyzed.
A Common Mistake: Assuming Every Stationary Point Is an Extremum
A common error is to see \(P'(r)=0\) and conclude that \(P\) must have a maximum or minimum there. The equation \(P'(r)=0\) only identifies a stationary point. It does not tell us whether the derivative changes sign.
For example, with \(P(x)=x^3-3x^2+3x\), a mistaken argument might stop at \(P'(1)=3(1-1)^2=0\) and call \(x=1\) a minimum. But the derivative is positive on both sides of \(1\), so the function increases through the point. The correction is to check the derivative’s sign on both sides: the zero has even multiplicity, and there is no extremum.
A related mistake is to assume that multiplicity alone tells us whether a point is a maximum or a minimum. Odd multiplicity guarantees a sign change, not its direction. The remaining factor’s sign determines the direction: inspect it, or test a point on each side.
A Reliable Classification Process
For a polynomial derivative that is factored near its real zeros, use the following sequence. It is the direction of the sign change—not just the existence of a derivative zero—that determines the classification.
Set \(P'(x)=0\) and identify the real solutions.
Write the derivative with its factors and count the power of each corresponding linear factor.
Even multiplicity means no change; odd multiplicity means a change.
Check the signs on both sides. A change from negative to positive gives a minimum; a change from positive to negative gives a maximum.
This method applies at isolated zeros of the derivative. A nonconstant polynomial has a nonzero polynomial as its derivative, and that derivative has only finitely many zeros. If \(P'\) is identically zero, then \(P\) is constant and there is no isolated derivative zero to classify by this multiplicity test.
Check Your Understanding
For each question, use the derivative’s factorization and, where needed, determine its signs on either side of the zero.
- If \(P'(x)=5(x-3)^4\), what is the multiplicity of the zero at \(3\), and does \(P'\) change sign there?
- If \(P'(x)=-(x+2)^3\), what are the signs of \(P'\) to the left and right of \(-2\), and what type of local extremum does \(P\) have there?
- Can a stationary point be a local extremum when its derivative zero has even multiplicity? Explain using the sign-change rule.
- For \(P'(x)=(x-1)^2(x+1)\), find the multiplicity of each zero and classify each stationary point.
- Why does the multiplicity determine whether the sign changes but not, by itself, whether the point is a maximum or a minimum?
