Rolle's Theorem

Equal Heights Force a Horizontal Tangent

Fermat’s Theorem says that a differentiable function has derivative zero at an interior local extremum. Rolle’s Theorem connects that local conclusion to information given at two endpoints: if a function takes the same value at both ends of an interval, then somewhere between them its derivative is zero. The function need not have an extremum at either endpoint. Instead, continuity on the entire closed interval ensures that its highest and lowest values are attained, and differentiability controls what happens at an interior extremum.

Rolle’s Theorem was stated in “Mean Value Theorem Preview.” We recall its conditions here because each one matters in applications. The theorem applies to a function continuous on a closed interval \([a,b]\), differentiable at every point of the open interval \((a,b)\), and satisfying \(f(a)=f(b)\). It guarantees at least one \(c\in(a,b)\) such that \(f'(c)=0\). The conclusion concerns an interior point, not either endpoint.

Rolle’s Theorem (recall): Let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be continuous on \([a,b]\) and differentiable on \((a,b)\). If \(f(a)=f(b)\), then there exists \(c\in(a,b)\) such that \(f'(c)=0\).

Geometrically, the graph begins and ends at the same height. Rolle’s Theorem says that at some intermediate point its tangent is horizontal. That point is not necessarily unique, and the theorem does not say that every point with derivative zero is a local extremum. Its value is more precise: equal endpoint values force at least one stationary point in between.

Reading the Conditions Carefully

The closed interval matters because a continuous function on \([a,b]\) attains an absolute maximum and an absolute minimum, by the Extreme Value Theorem. If the function is constant, every interior point has derivative zero. If it is not constant, at least one of its absolute extrema must occur strictly between the endpoints: the endpoint values agree, so an interior value that is larger or smaller than that common value gives an interior extremum. Fermat’s Theorem then supplies a point with derivative zero. This is the central idea behind Rolle’s Theorem; its full proof is the subject of the next tutorial.

Continuity is required at the endpoints as well as in the interior. Differentiability is required only in the open interval, since the point \(c\) is guaranteed to lie there. Equal endpoint values are also essential: without them, even a straight line can have no horizontal tangent. These are not formal details to be checked after the calculation; they are what make the conclusion valid.

Worked Examples: Locating the Guaranteed Point

Worked Example: A Quadratic with Equal Endpoint Values

Let \(f(x)=x^2-4x+1\) on \([0,4]\). This polynomial is continuous on the closed interval and differentiable on its interior. Its endpoint values are

f(0)=0^2-4(0)+1=1, \qquad f(4)=4^2-4(4)+1=16-16+1=1.

Rolle’s Theorem therefore guarantees a \(c\in(0,4)\) with \(f'(c)=0\). Since \(f'(x)=2x-4\), the equation is \(2c-4=0\), so \(c=2\). Indeed, the graph descends from height \(1\), reaches its minimum at \(x=2\), and returns to height \(1\).

Worked Example: More Than One Stationary Point

Consider \(g(x)=x^3-3x\) on \([-\sqrt{3},\sqrt{3}]\). At the right endpoint,

g(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3} =3\sqrt{3}-3\sqrt{3}=0.

At the left endpoint,

g(-\sqrt{3})=(-\sqrt{3})^3-3(-\sqrt{3}) =-3\sqrt{3}+3\sqrt{3}=0.

The hypotheses hold, so Rolle’s Theorem guarantees a stationary point. Here the derivative is \(g'(x)=3x^2-3\), which vanishes at \(x=-1\) and \(x=1\); both lie in \((-\sqrt{3},\sqrt{3})\). The theorem promises at least one such point, not exactly one.

Worked Example: An Endpoint Condition That Fails

Let \(h(x)=2x+1\) on \([0,3]\). The function is continuous and differentiable, but its endpoint values are

h(0)=1, \qquad h(3)=7.

Because these values are unequal, Rolle’s Theorem does not apply. In fact, \(h'(x)=2\) at every point, so there is no \(c\in(0,3)\) with \(h'(c)=0\). This example shows why equal endpoint values are a genuine hypothesis rather than a convenient extra condition.

A Positive Derivative Gives Strict Increase

Rolle’s Theorem can also compare values at different points. If a differentiable function had the same value at two distinct points, Rolle’s Theorem would force its derivative to vanish somewhere between them. Thus, when the derivative is positive everywhere, equal values are impossible. To determine the order of the values, we apply Rolle’s Theorem to a function formed by subtracting the line joining two points.

Theorem (Positive Derivative Implies Strict Increase): Let \(I\) be an interval, and suppose \(f:I\to\mathbb{R}\) is continuous on \(I\) and differentiable at every interior point of \(I\). If \(f'(t)>0\) at every interior point \(t\in I\), then \(f\) is strictly increasing on \(I\): whenever \(x,y\in I\) and \(x<y\), we have \(f(x)<f(y)\).

Proof. Fix \(x,y\in I\) with \(x<y\). The entire segment \([x,y]\) lies in \(I\), since \(I\) is an interval. Define the slope between the endpoint values by

m=\frac{f(y)-f(x)}{y-x},

and define \(g(t)=f(t)-m(t-x)\) for \(t\in[x,y]\). The function \(g\) is continuous on \([x,y]\) and differentiable on \((x,y)\). Also,

g(x)=f(x), \qquad g(y)=f(y)-m(y-x)=f(y)-\bigl(f(y)-f(x)\bigr)=f(x).

Rolle’s Theorem gives some \(c\in(x,y)\) such that \(g'(c)=0\). Since \(g'(t)=f'(t)-m\), this means \(m=f'(c)>0\). Because \(y-x>0\), the equation \(f(y)-f(x)=m(y-x)\) implies \(f(y)-f(x)>0\). Therefore \(f(x)<f(y)\), as required. \(\square\)

Worked Example: Proving a Function Has at Most One Zero

Let \(q(x)=x^3+2x\) on \(\mathbb{R}\). Its derivative is

$$ q'(x)=3x^2+2>0 $$

for every real \(x\), because \(x^2\geq0\). The theorem just proved shows that \(q\) is strictly increasing. It can therefore have at most one zero: if \(q(u)=q(v)=0\) for \(u<v\), strict increase would give \(q(u)<q(v)\), a contradiction. In this example \(q(x)=x(x^2+2)\), and \(x^2+2>0\) for every real \(x\), so its only zero is \(x=0\).

Counting Zeros by Counting Derivative Zeros

A second useful consequence is a bound on how many distinct zeros a function can have. Between every two distinct zeros of a continuous function, Rolle’s Theorem produces a zero of the derivative. The intervals between consecutive zeros do not overlap in their interiors, so the derivative zeros produced in this way are distinct. This lets us turn information about \(f'\) into a limit on the number of zeros of \(f\).

Theorem (Zero-Count Bound from Rolle’s Theorem): Let \(I\) be an interval, let \(n\) be a nonnegative integer, and suppose \(f:I\to\mathbb{R}\) is continuous on \(I\) and differentiable at every interior point of \(I\). If \(f'\) has at most \(n\) distinct zeros in the interior of \(I\), then \(f\) has at most \(n+1\) distinct zeros in \(I\).

Proof. Suppose, to the contrary, that \(f\) has at least \(n+2\) distinct zeros in \(I\). Choose \(n+2\) of them and arrange them in increasing order:

x_1<x_2<\cdots<x_{n+2}, \qquad f(x_i)=0 \quad (1\leq i\leq n+2).

For each \(i=1,\ldots,n+1\), the function is continuous on \([x_i,x_{i+1}]\), differentiable on \((x_i,x_{i+1})\), and has equal values at the two endpoints, since both values are zero. Rolle’s Theorem supplies a point \(c_i\in(x_i,x_{i+1})\) with \(f'(c_i)=0\). These points are distinct: if \(i<j\), then \(c_i<x_{i+1}\leq x_j<c_j\). Thus \(f'\) has at least \(n+1\) distinct zeros, contradicting the assumption that it has at most \(n\). Therefore \(f\) has at most \(n+1\) distinct zeros. \(\square\)

Worked Example: A Quartic with Four Distinct Zeros

Consider \(p(x)=x^4-5x^2+4\). Its derivative factors as

$$ p'(x)=4x^3-10x=2x(2x^2-5). $$

Thus \(p'\) has exactly three distinct real zeros: \(0\) and \(x=\pm\sqrt{5/2}\). The zero-count bound says that \(p\) can have at most four distinct real zeros. Factoring the polynomial gives

$$ p(x)=(x^2-1)(x^2-4). $$

So its zeros are \(x=-2,-1,1,2\), all distinct. Direct substitution checks the factorization’s roots: at \(x=\pm1\), \(x^2-1=0\); at \(x=\pm2\), \(x^2-4=0\). The bound is attained exactly.

What Rolle’s Theorem Does—and Does Not—Say

Rolle’s Theorem is useful whenever equality at two points can be arranged. It can rule out repeated function values when the derivative has a fixed nonzero sign, as in the strict-increase result. It can also control the number of solutions to an equation: apply the zero-count bound to \(f(x)-L\) to count the points where \(f(x)=L\), provided the derivative has a known number of zeros.

A common error is to conclude that \(f'(c)=0\) from \(f(a)=f(b)\) without checking the other hypotheses. For example, equal endpoint values alone are not enough if the function is discontinuous inside the interval or fails to be differentiable at an interior point. Another mistake is to expect exactly one stationary point: Rolle’s Theorem guarantees existence, and the cubic example had two. Finally, the theorem does not claim the stationary point is a maximum or minimum; a derivative can vanish at a point that is neither.

The main habit to develop is to identify the interval on which the theorem will be applied, verify continuity on the closed segment and differentiability on its interior, and check that the two endpoint values agree. Once those facts are in place, Rolle’s Theorem converts equality of values into a zero of the derivative. The next tutorial proves that conversion in detail.

Check Your Understanding

Use the hypotheses and consequences of Rolle’s Theorem to answer the following.

State the continuity, differentiability, and endpoint-value requirements in Rolle’s Theorem.
For \(r(x)=x^2-6x+5\) on \([1,5]\), verify the endpoint values and find a point guaranteed by Rolle’s Theorem.
Why does \(f'(x)>0\) throughout an interval prevent \(f\) from taking the same value at two distinct points?
If \(f'\) has at most two distinct zeros on an interval, what is the largest possible number of distinct zeros of \(f\), under the zero-count theorem’s hypotheses?
Does Rolle’s Theorem guarantee exactly one point where the derivative is zero? Explain briefly.

What You'll Learn