Two Ways Roots Lead to Derivative Roots
A function can have a derivative root for more than one reason. If a differentiable function has the same value at two distinct points, Rolle’s Theorem guarantees a point between them where the derivative is zero. For a polynomial, a repeated root also forces the derivative to vanish at that root itself.
These are related but different conclusions. Rolle’s Theorem concerns an interval whose endpoints are distinct roots. The repeated-root conclusion comes from factoring the polynomial and differentiating. Keeping the two arguments separate helps identify exactly where a derivative root is guaranteed and why.
Rolle’s Theorem Between Distinct Roots
Rolle’s Theorem applies to a function \(f\) on a closed interval \([a,b]\) when \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and has equal endpoint values \(f(a)=f(b)\). It then guarantees at least one \(c\in(a,b)\) such that \(f'(c)=0\).
In particular, if \(a\) and \(b\) are distinct roots of a function satisfying those continuity and differentiability conditions, then \(f(a)=f(b)=0\), so \(f(a)=f(b)\). Rolle’s Theorem applies. The point \(c\) it guarantees lies strictly between the roots; it is not either endpoint.
Polynomials satisfy the required continuity and differentiability conditions on every closed interval, so this result applies to any two distinct real roots of a polynomial. The theorem guarantees that a derivative root exists, but it does not generally identify its exact location.
Worked Example: Derivative Roots Between Three Distinct Roots
Consider \(p(x)=(x+2)x(x-2)\). Its three distinct roots are \(-2\), \(0\), and \(2\). Because \(p\) is a polynomial, it is continuous and differentiable everywhere.
First expand the polynomial:
Differentiate:
Apply Rolle’s Theorem on \([-2,0]\). The endpoint values are \(p(-2)=0\) and \(p(0)=0\), so there is a derivative root in \((-2,0)\). On \([0,2]\), \(p(0)=0\) and \(p(2)=0\), so there is a derivative root in \((0,2)\).
In this example, solve for the derivative roots to check the locations explicitly:
Since \(1<\frac{2}{\sqrt{3}}<2\), the positive derivative root lies in \((0,2)\), and the negative one lies in \((-2,0)\). Finally, substituting into the derivative gives
Thus the calculated roots agree with Rolle’s guarantee on both intervals.
Multiplicity of a Polynomial Root
To describe a repeated root precisely, use the factorization of a polynomial. A real number \(r\) is a root of multiplicity \(m\) of a polynomial \(p\) if
for some polynomial \(q\) and positive integer \(m\). The condition \(q(r)\ne0\) ensures that the displayed power gives the full multiplicity, rather than just part of it. A root is repeated if its multiplicity is at least \(2\).
For example, if \(p(x)=(x-3)^2(x+1)\), then \(3\) is a root of multiplicity \(2\): the factor \((x-3)\) occurs twice, and the remaining factor has value \(4\) at \(x=3\), which is nonzero.
Why a Repeated Root Is Also a Derivative Root
The factorization defining multiplicity gives a direct explanation. Differentiate \(p(x)=(x-r)^m q(x)\) using the product rule:
If \(m\ge2\), then \((x-r)^{m-1}\) contains at least one factor of \(x-r\). Consequently, \(p'(r)=0\). More precisely, the factor in parentheses has value \(mq(r)\) at \(r\), and \(mq(r)\ne0\). So that factor does not add another factor of \(x-r\): the multiplicity of \(r\) as a root of \(p'\) is exactly \(m-1\).
This conclusion concerns the same point \(r\), unlike Rolle’s Theorem, which guarantees a derivative root strictly between two distinct roots. For a simple root, \(m=1\), the factorization argument does not force \(p'(r)=0\): the derivative need not vanish there.
Worked Example: Differentiating a Polynomial with Repeated Roots
Let \(p(x)=(x-1)^3(x+2)^2\). The factorization shows that \(1\) has multiplicity \(3\) and \(-2\) has multiplicity \(2\). Differentiate using the product rule:
At \(x=1\), the derivative has a factor \((x-1)^2\). The other factors have values \(3\) and \(9\), both nonzero, so \(1\) has multiplicity exactly \(2\) as a root of \(p'\), one less than its multiplicity in \(p\). At \(x=-2\), the derivative has a factor \(x+2\); the other factors have values \(9\) and \(-6\), both nonzero. Thus \(-2\) has multiplicity exactly \(1\) as a root of \(p'\).
The two distinct roots \(-2\) and \(1\) of \(p\) also trigger Rolle’s Theorem on \([-2,1]\). The derivative root \(-\frac45\), obtained from \(5x+4=0\), lies strictly between them. Check it in the derivative:
The derivative therefore has roots at both original repeated-root locations and at a point between the two distinct roots. The factorization displays these roots explicitly; Rolle’s Theorem independently guarantees an interior one.
A Second Example: Separating the Two Conclusions
Worked Example: A Repeated Root and an Interior Derivative Root
Take \(p(x)=(x+2)(x-1)^2\). The roots of \(p\) are \(-2\) and \(1\), with \(1\) repeated. By the product rule,
The repeated root \(1\) is a root of the derivative: substituting gives \(p'(1)=3(0)(2)=0\). The distinct roots \(-2\) and \(1\) give a separate Rolle conclusion on \([-2,1]\). Indeed, the explicit derivative root \(x=-1\) lies in \((-2,1)\), and
Also check the original endpoint values: \(p(-2)=0\) because of the factor \(x+2\), and \(p(1)=0\) because of the factor \((x-1)^2\). Thus the interval satisfies Rolle’s equal-endpoint condition, while the derivative root at \(1\) follows from the repeated factor.
Common Pitfall: Confusing the Interval Point with a Repeated Root
A common error is to assume that Rolle’s Theorem itself says a derivative vanishes at a repeated root. Its conclusion is that \(f'(c)=0\) for some \(c\) strictly inside an interval \((a,b)\) whose endpoints are distinct. It does not say that \(f'\) vanishes at either endpoint.
For example, one might write, incorrectly, “The repeated root \(x=1\) is the endpoint of an interval with itself, so Rolle’s Theorem gives \(p'(1)=0\).” The interval \([1,1]\) has no distinct endpoints and no interior point, so this is not a valid application. The correction is to use the polynomial factorization. If \(p(x)=(x-1)^2q(x)\), with \(q(1)\ne0\), then
By contrast, if a polynomial has two distinct roots \(a<b\), apply Rolle’s Theorem on \([a,b]\) to obtain a derivative root in \((a,b)\). These are different arguments, and each has its own conditions and conclusion.
Why the Distinction Matters
Knowing why a derivative root is guaranteed helps avoid claiming more than the hypotheses support. Rolle’s Theorem is an existence result: it establishes an interior point but may not locate it. Factorization gives more specific information for polynomial repeated roots, including the exact multiplicity remaining after differentiation.
When analyzing a polynomial, first identify which roots are distinct and which are repeated. Use Rolle’s Theorem between distinct roots, checking continuity and differentiability on the relevant interval. Use factorization and the product rule at a repeated root. In either case, verify an explicit derivative root by substitution when the derivative has been calculated.
Check Your Understanding
Answer each question by identifying the relevant theorem or factorization argument.
- If a differentiable function is continuous on \([a,b]\), and \(a<b\) with \(f(a)=f(b)=0\), where does Rolle’s Theorem guarantee a point \(c\) with \(f'(c)=0\)?
- For a polynomial root \(r\) of multiplicity \(m\ge2\), what is its multiplicity as a root of the derivative?
- Why does the definition \(p(x)=(x-r)^m q(x)\) require \(q(r)\ne0\)?
- For \(p(x)=(x-4)^3(x+1)\), what is the multiplicity of \(4\) as a root of \(p'\)?
- Does Rolle’s Theorem alone guarantee that the derivative vanishes at either endpoint of the interval? Explain briefly.
