Why Multiplicity Matters
A polynomial derivative can equal zero at a point without the original function having a local maximum or minimum there. The key question is whether the derivative changes sign as it passes through zero. For a polynomial, the multiplicity of the derivative’s root provides a reliable way to answer that question.
This gives a useful refinement of derivative sign charts: instead of testing many values near a root, we can often infer the sign behavior directly from the factorization. The method applies to a polynomial function \(f\) near a stationary point \(x=r\), where \(f'(r)=0\).
Multiplicity of a Derivative Root
A root \(r\) of a polynomial \(p\) has multiplicity \(m\) if \(p(x)\) contains the factor \((x-r)^m\), but does not contain \((x-r)^{m+1}\). Equivalently, near that root we can write \(p(x)=(x-r)^m g(x)\), where \(g\) is a polynomial and \(g(r)\ne 0\).
For this tutorial, \(p\) will be the derivative \(f'\). The multiplicity in question is therefore the multiplicity of \(r\) as a root of \(f'\), not the multiplicity of \(r\) as a root of \(f\). These are different questions: for instance, \(f(x)=x^3\) has a root of multiplicity three at zero, while \(f'(x)=3x^2\) has a root of multiplicity two there.
The Sign Rule for Polynomial Derivatives
Suppose \(f'\) has a root of multiplicity \(m\) at \(r\). Then we can factor the derivative as
Because \(g\) is continuous and \(g(r)\ne0\), it keeps the same sign on a sufficiently small interval on either side of \(r\). Thus, near \(r\), the sign change of \(f'\) is determined by \((x-r)^m\). If \(m\) is even, that factor is positive on both sides. If \(m\) is odd, it has opposite signs on the two sides.
- If \(m\) is odd, \(f'\) changes sign at \(r\). The function has a local extremum there: a local minimum if \(f'\) changes from negative to positive, and a local maximum if it changes from positive to negative.
- If \(m\) is even, \(f'\) does not change sign at \(r\). The function has no local extremum there. For a polynomial \(f\), the point is a stationary inflection point.
The extrema conclusion follows from the meaning of derivative signs: a function decreases where its derivative is negative and increases where its derivative is positive. For the even-multiplicity case, the inflection conclusion needs one additional check: the concavity changes. The derivation below verifies that conclusion as well as the sign rule.
Why the Sign Rule Works
Here is a complete local argument. Write \(f'(x)=(x-r)^m g(x)\), with \(g(r)\ne0\). Since \(g\) is continuous, there is an interval around \(r\) on which \(g\) has the same sign as \(g(r)\). It cannot introduce a sign change inside that interval.
For \(x<r\), \(x-r<0\); for \(x>r\), \(x-r>0\). Raising this factor to an even power makes it positive on both sides, while an odd power preserves its opposite signs.
Since \(g\) keeps one sign near \(r\), \(f'\) has no sign change when \(m\) is even and does change sign when \(m\) is odd.
An odd multiplicity gives either decrease then increase, a local minimum, or increase then decrease, a local maximum. An even multiplicity gives the same direction of motion on both sides, so there is no local extremum.
To confirm the inflection claim for even \(m\), differentiate the factorization:
The expression in parentheses is continuous and at \(x=r\) equals \(m g(r)\), which is nonzero. It therefore keeps one sign sufficiently near \(r\). When \(m\) is even, \(m-1\) is odd, so \((x-r)^{m-1}\) changes sign at \(r\). Consequently \(f''\) changes sign there, and the concavity changes. Also \(f'(r)=0\), so the point is stationary: it is a stationary inflection point.
Worked Examples
Worked Example: An Odd-Multiplicity Root Gives a Local Minimum
Classify the stationary point of \(f(x)=(x-1)^4\).
Differentiate:
The derivative is zero at \(x=1\), and its root there has multiplicity \(3\), which is odd. To determine which extremum occurs, check the sign on both sides. At \(x=0\), \(f'(0)=4(-1)^3=-4<0\). At \(x=2\), \(f'(2)=4(1)^3=4>0\). Thus \(f'\) changes from negative to positive, so \(f\) decreases and then increases: \(x=1\) is a local minimum.
At the point, \(f(1)=(1-1)^4=0\). A value check also illustrates the nearby behavior: \(f(0)=1\) and \(f(2)=1\), both greater than \(f(1)\). Finally, differentiating the proposed derivative verifies the calculation: \(\frac{d}{dx}(x-1)^4=4(x-1)^3\).
Worked Example: An Even-Multiplicity Root Gives a Stationary Inflection Point
Classify the stationary point of \(f(x)=(x+2)^3\).
Differentiate:
The derivative root \(x=-2\) has multiplicity \(2\), which is even. For \(x=-3\), \(f'(-3)=3(-1)^2=3>0\). For \(x=-1\), \(f'(-1)=3(1)^2=3>0\). The derivative is positive on both sides, so there is no change from increasing to decreasing or from decreasing to increasing. Therefore \(x=-2\) is not a local extremum.
To verify the inflection classification, differentiate again: \(f''(x)=6(x+2)\). For \(x=-3\), \(f''(-3)=-6<0\); for \(x=-1\), \(f''(-1)=6>0\). Concavity changes, and \(f'(-2)=0\), so \(x=-2\) is a stationary inflection point. The function value is \(f(-2)=0\). Checking the derivative by differentiation gives \(\frac{d}{dx}(x+2)^3=3(x+2)^2\), as used.
Worked Example: Classifying Two Different Derivative Roots
Find and classify the stationary points of \(f(x)=\frac{x^4}{4}-\frac{3x^2}{2}+2x\).
Differentiate and factor:
The factorization shows stationary points at \(x=-2\) and \(x=1\). At \(x=-2\), the root has multiplicity \(1\), which is odd. For \(x=-3\), \(f'(-3)=(-4)^2(-1)=-16<0\); for \(x=0\), \(f'(0)=(-1)^2(2)=2>0\). The derivative changes from negative to positive, so \(x=-2\) is a local minimum. Its function value is
At \(x=1\), the root has multiplicity \(2\), which is even. For \(x=0\), \(f'(0)=2>0\); for \(x=2\), \(f'(2)=(1)^2(4)=4>0\). The derivative has the same sign on both sides, so \(x=1\) is not an extremum. Its concavity changes because
At \(x=0\), \(f''(0)=-3<0\); at \(x=2\), \(f''(2)=9>0\). Thus \(x=1\) is a stationary inflection point, with \(f(1)=\frac14-\frac32+2=\frac34\). To check the original differentiation, differentiating the three terms gives \(x^3-3x+2\), and expanding \((x-1)^2(x+2)=(x^2-2x+1)(x+2)=x^3-3x+2\), confirming the factorization.
A Common Mistake
A frequent error is to conclude that every point where \(f'(x)=0\) must be a local extremum. For \(f(x)=x^3\), that mistaken reasoning might say: “\(f'(0)=0\), so \(x=0\) is a local minimum.” The conclusion does not follow from the equation \(f'(0)=0\) alone.
In fact, \(f'(x)=3x^2\). At \(x=-1\), \(f'(-1)=3>0\), and at \(x=1\), \(f'(1)=3>0\). The derivative does not change sign at zero, so the function is increasing on both sides and has no local extremum there. Since \(f''(x)=6x\) changes from negative to positive at zero, \((0,0)\) is a stationary inflection point.
How to Use the Rule Reliably
For a polynomial, factor \(f'\) and identify the multiplicity of each root. An odd multiplicity means the derivative changes sign, so the point is a local extremum; the direction of the sign change distinguishes maximum from minimum. An even multiplicity means no derivative sign change and, for a polynomial, a stationary inflection point. If a factorization is not available or a sign is uncertain, test one value on each side and make a sign chart.
The rule is local: it describes what happens sufficiently close to the root. Other derivative roots may lie elsewhere and change the function’s behavior outside that neighborhood. Keeping the classification local prevents a sign conclusion near one point from being applied to an entire domain without justification.
Check Your Understanding
For each question, use the multiplicity of the root of the derivative and, when needed, verify signs on either side.
- If \(f'(x)=(x-3)^4\), does \(f\) have a local extremum at \(x=3\)? What type of stationary point is it?
- If \(f'(x)=-(x+1)^5\), classify the stationary point at \(x=-1\) as a local maximum or minimum.
- For \(f(x)=x^5\), what is the multiplicity of \(0\) as a root of \(f'\), and what happens to the derivative’s sign there?
- Why does an even-multiplicity root of a polynomial derivative give a change in concavity?
- Can the fact that \(f'(r)=0\), by itself, establish that \(f\) has a local extremum at \(r\)? Explain.
