Rolle's theorem

Theorem

Rolle's theorem

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Statement

Let \( a, b \in \mathbb{R} \) with \( a \lt b \), and let \( f \colon [a,b] \to \mathbb{R} \) be a function satisfying all three of: (i) \( f \) is continuous on the closed interval \( [a,b] \); (ii) \( f \) is differentiable on the open interval \( (a,b) \); (iii) \( f(a) = f(b) \). Then there exists at least one point \( c \in (a,b) \) with \[ f'(c) = 0. \] Nothing is claimed about uniqueness of \( c \), and differentiability is not required at the endpoints \( a \) and \( b \).

Why it matters

Rolle's theorem is the hinge between topology and differential calculus: it takes a purely topological fact (a continuous function on a compact interval attains its extrema) and converts it into a differential fact (the derivative vanishes somewhere). Almost every "there exists a point where the derivative does X" statement in a first analysis course is Rolle in disguise: the Mean Value Theorem is Rolle after shearing by a linear function, Cauchy's Mean Value Theorem is Rolle applied to a cleverly built auxiliary function, and the Lagrange remainder in Taylor's theorem comes from applying Rolle repeatedly.

It is also the standard engine for root counting: between any two zeros of a differentiable function lies a zero of its derivative, so a polynomial or transcendental equation can be shown to have at most \( k \) roots by controlling the zeros of derivatives. Paired with the Intermediate Value Theorem (which gives at least), this yields exact root counts — a technique used constantly in analysis, numerical analysis, and approximation theory.

Hypotheses

Hypothesis (continuity on the closed interval \( [a,b] \)).Drop continuity at even one endpoint and the conclusion dies: define \( f(x) = x \) for \( 0 \le x \lt 1 \) and \( f(1) = 0 \). Then \( f(0) = f(1) = 0 \), \( f \) is differentiable on \( (0,1) \), yet \( f'(x) = 1 \) for every \( x \in (0,1) \). The maximum is "attained in the limit" at \( x = 1 \) but the jump destroys it.

Hypothesis (differentiability on the open interval \( (a,b) \)).Take \( f(x) = |x| \) on \( [-1,1] \): continuous everywhere, \( f(-1) = f(1) = 1 \), but \( f'(x) = \pm 1 \) wherever the derivative exists. A single interior point of non-differentiability (the corner at \( 0 \), exactly where the minimum sits) is enough to break the theorem.

Hypothesis (equal endpoint values \( f(a) = f(b) \)).With \( f(x) = x \) on \( [0,1] \) every other hypothesis holds but \( f'(x) = 1 \) everywhere. Unequal endpoint values shear the picture; the correct statement then is the Mean Value Theorem, not Rolle.

Hypothesis (the domain is a real interval — completeness is load-bearing).Work over \( \mathbb{Q} \) instead of \( \mathbb{R} \): the function \( f(x) = (x^2 - 2)^2 \) restricted to \( \mathbb{Q} \cap [0,2] \) is a polynomial, hence "continuous and differentiable" in the \( \mathbb{Q} \)-limit sense, and \( f(0) = f(2) = 4 \); but \( f'(x) = 4x(x^2 - 2) \) vanishes only at \( x = 0 \) and \( x = \pm\sqrt{2} \), none of which is an interior rational point. Rolle secretly invokes the completeness of \( \mathbb{R} \) through the Extreme Value Theorem.

Hypothesis (real-valued codomain).For \( f \colon [0, 2\pi] \to \mathbb{C} \), \( f(t) = e^{it} \), we have \( f(0) = f(2\pi) = 1 \) and \( f \) smooth, yet \( |f'(t)| = |ie^{it}| = 1 \) for all \( t \): the derivative never vanishes. Rolle uses the order structure of \( \mathbb{R} \) (max and min), which \( \mathbb{C} \) and \( \mathbb{R}^n \) lack.

Proof

We use two named ingredients: the Extreme Value Theorem (Weierstrass: a continuous real-valued function on a compact interval attains a maximum and a minimum), which we cite; and Fermat's interior extremum lemma, which we do not cite but prove in full as Steps 4–7.

\( f \) is continuous on \( [a,b] \), which is closed and bounded, so there exist \( p, q \in [a,b] \) with \[ f(p) = M := \sup_{x \in [a,b]} f(x), \qquad f(q) = m := \inf_{x \in [a,b]} f(x), \] and \( m \le f(x) \le M \) for all \( x \in [a,b] \).

Extreme Value Theorem (Weierstrass), applicable because \( [a,b] \) is compact and \( f \) is continuous on all of it — this is where continuity at the endpoints is consumed. A

Case 1: \( M = m \). Then \( f \) is constant on \( [a,b] \), so \( f'(c) = 0 \) for every \( c \in (a,b) \); since \( a \lt b \), the open interval \( (a,b) \) is nonempty and any such \( c \) witnesses the conclusion.

If \( \inf = \sup \) then \( f(x) = M \) for all \( x \); the derivative of a constant is \( 0 \) directly from the difference-quotient definition. The hypothesis \( a \lt b \) guarantees a witness exists. A

Case 2: \( m \lt M \). Then at least one of the extremum points \( p, q \) lies in the open interval \( (a,b) \).

Key idea. Suppose not: then \( p, q \in \{a, b\} \), so \( M = f(p) \) and \( m = f(q) \) both belong to \( \{f(a), f(b)\} \). But hypothesis (iii) gives \( f(a) = f(b) \), so that set is a single value and \( M = m \), contradicting \( m \lt M \). This is precisely where the equal-endpoint hypothesis forces an extremum inward. B

Without loss of generality the maximum is interior: \( c := p \in (a,b) \) and \( f(x) \le f(c) \) for all \( x \in [a,b] \). (If instead only the minimum is interior, replace \( f \) by \( -f \); this negates the derivative and preserves every hypothesis, so a zero of \( (-f)' \) is a zero of \( f' \).)

Legitimate reduction: \( x \mapsto -f(x) \) is continuous on \( [a,b] \), differentiable on \( (a,b) \), has equal endpoint values, and swaps max with min. A

For \( x \in (c, b] \): \( \; x - c \gt 0 \) and \( f(x) - f(c) \le 0 \), hence \[ \frac{f(x) - f(c)}{x - c} \le 0 \quad \text{for all } x \in (c, b]. \] Since \( c \in (a,b) \), \( f \) is differentiable at \( c \), so the two-sided limit of the difference quotient exists and equals \( f'(c) \); in particular the right-hand limit exists and \[ f'(c) = \lim_{x \to c^{+}} \frac{f(x) - f(c)}{x - c} \le 0. \]

The quotient inequality is arithmetic from \( f(c) = M \). Passing to the limit preserves non-strict inequalities: this is the order-limit theorem for function limits (if \( g(x) \le 0 \) on a punctured right neighbourhood and \( \lim_{x \to c^{+}} g(x) = L \) exists, then \( L \le 0 \); proof by contradiction with \( \varepsilon = L/2 \)). C

For \( x \in [a, c) \): \( \; x - c \lt 0 \) and \( f(x) - f(c) \le 0 \), hence \[ \frac{f(x) - f(c)}{x - c} \ge 0 \quad \text{for all } x \in [a, c), \] and by the same order-limit argument on the left, \[ f'(c) = \lim_{x \to c^{-}} \frac{f(x) - f(c)}{x - c} \ge 0. \]

Dividing the non-positive numerator by the negative quantity \( x - c \) flips the sign. Again the limit exists because \( f \) is differentiable at the interior point \( c \), and non-strict inequalities survive limits. C

\[ 0 \le f'(c) \le 0 \implies f'(c) = 0. \]

Antisymmetry of \( \le \) on \( \mathbb{R} \). Steps 4–7 constitute a complete proof of Fermat's interior extremum lemma; combined with Cases 1 and 2 the theorem is proved in all cases. \( \blacksquare \) A

Result

\[ f \in C[a,b], \;\; f \text{ differentiable on } (a,b), \;\; f(a) = f(b) \;\; \Longrightarrow \;\; \exists\, c \in (a,b): \; f'(c) = 0. \]

Reading. If a smooth-enough graph starts and ends at the same height, it must have a horizontal tangent strictly between the endpoints. You cannot return to your starting altitude without, at some instant, moving neither up nor down.

Scope. Real-valued functions of one real variable on a compact interval \( [a,b] \) with \( a \lt b \). Continuity is needed on all of \( [a,b] \) (endpoints included); differentiability only on \( (a,b) \). The conclusion fails in general for complex- or vector-valued functions, and over incomplete ordered fields such as \( \mathbb{Q} \). The point \( c \) exists but is not constructed, located, or unique.

Corollaries & converses

Mean Value Theorem. Apply Rolle to \( g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a) \), which satisfies \( g(a) = g(b) = 0 \); the resulting \( c \) gives \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Rolle is exactly the MVT normalised to zero slope.
Zeros interlace. Between any two zeros of a differentiable function lies a zero of its derivative. Iterating: if \( f \) is \( n \) times differentiable with \( n + 1 \) distinct zeros, then \( f^{(n)} \) has a zero.
Root bounds. A polynomial of degree \( n \) whose derivative has \( k \) real roots has at most \( k + 1 \) real roots; more generally, if \( f' \) never vanishes on an interval then \( f \) is injective there (contrapositive of Rolle).
Cauchy MVT and Taylor's remainder. Both follow by applying Rolle to suitable auxiliary functions; the Lagrange form \( \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1} \) of the Taylor remainder comes from a single Rolle application to a function engineered to vanish at both \( a \) and \( x \).
Converse is FALSE. The converse would say: if \( f'(c) = 0 \) for some interior \( c \), then \( f \) takes equal values at two points surrounding it. Counterexample: \( f(x) = x^3 \) on \( [-1,1] \) has \( f'(0) = 0 \), yet \( f \) is strictly increasing, so no two points share a value. A stationary point does not imply a repeated height.

Fails without

No continuity at an endpoint: \( f(x) = x \) on \( [0,1) \), \( f(1) = 0 \). All other hypotheses hold, \( f(0) = f(1) \), yet \( f'(x) = 1 \) for every \( x \in (0,1) \). One removable-looking jump at a single endpoint kills the conclusion.
No differentiability at one interior point: \( f(x) = |x| \) on \( [-1,1] \). Continuous, \( f(-1) = f(1) = 1 \), but \( f'(x) = -1 \) on \( (-1,0) \) and \( f'(x) = 1 \) on \( (0,1) \): the derivative never vanishes. The extremum hides exactly at the corner.
No equal endpoint values: \( f(x) = x \) on \( [0,1] \): \( f' \equiv 1 \). (The repaired statement is the MVT.)
No completeness of the field: \( f(x) = (x^2 - 2)^2 \) on \( \mathbb{Q} \cap [0,2] \): \( f(0) = f(2) = 4 \), but \( f'(x) = 4x(x^2 - 2) \ne 0 \) at every interior rational, since that would need \( x = 0 \) or \( x^2 = 2 \).

Common errors

Demanding endpoint differentiability. Rolle needs differentiability only on the open interval. \( f(x) = \sqrt{1 - x^2} \) on \( [-1,1] \) has vertical tangents at \( \pm 1 \) and Rolle still applies (giving \( c = 0 \)). Rejecting such examples "because \( f'(\pm 1) \) doesn't exist" is wrong.
Checking differentiability but not endpoint continuity. Students verify \( f' \) exists on \( (a,b) \) and forget that continuity at \( a \) and \( b \) themselves is a separate, essential check (see the jump counterexample above).
Claiming uniqueness of \( c \). The theorem says "at least one". \( f(x) = \sin(2x) \) on \( [0, 2\pi] \) yields four such points. Writing "the point \( c \)" and then reasoning as if it is unique is a logic error.
Using Rolle when \( f(a) \ne f(b) \). That situation calls for the Mean Value Theorem; quoting Rolle there is a hypothesis violation even if the intended conclusion is salvageable.
Reading \( f'(c) = 0 \) as "c is a local extremum". The proof finds \( c \) at an extremum, but the boxed conclusion is only that the derivative vanishes — and conversely, in other problems a vanishing derivative need not mark an extremum (\( x^3 \) at \( 0 \)).
Treating the theorem as constructive. Rolle asserts existence; it gives no algorithm for locating \( c \). Arguments that need a specific numerical \( c \) must find it by other means (solving \( f'(x) = 0 \) directly, or numerically).

Discussion

Structurally, Rolle's theorem is a two-stage machine. Stage one is topological: compactness of \( [a,b] \) plus continuity forces \( f \) to attain its extrema (Weierstrass). Stage two is differential: an extremum trapped in the open interior forces the derivative to vanish there (Fermat). The single hypothesis \( f(a) = f(b) \) exists purely to guarantee that at least one extremum is trapped in the interior rather than escaping to the boundary. Seen this way, the theorem is the minimal honest bridge between "compactness gives extrema" and "extrema give critical points", and every one of its hypotheses is consumed at an identifiable step of the proof — a model worth internalising for how tightly stated theorems work.

Historically, Michel Rolle published a version of the result in 1691 — for polynomials only, and by algebraic means (the "method of cascades", essentially repeated formal differentiation, building on Hudde's rule) rather than by any limiting argument. The irony is famous: Rolle was one of the loudest contemporary critics of the infinitesimal calculus, dismissing it as a "collection of ingenious fallacies", yet his name is now welded to one of its foundational lemmas. The modern statement and proof, resting on Weierstrass's Extreme Value Theorem, had to wait for the nineteenth-century rigourisation of analysis; the name "Rolle's theorem" was attached in that era (Drobisch in German, Bellavitis in Italian usage).

The theorem's descendants dominate a first analysis course: the Mean Value Theorem, Cauchy's Mean Value Theorem, L'Hôpital-type arguments, the Lagrange and Cauchy remainder forms of Taylor's theorem, and the error term \( \frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^{n}(x - x_i) \) in polynomial interpolation — that last one obtained by applying Rolle \( n + 1 \) times in succession to an auxiliary function with \( n + 2 \) zeros. In the algebraic direction, "between two roots of \( f \) lies a root of \( f' \)" is the real-analytic ancestor of results like the Gauss–Lucas theorem (roots of \( p' \) lie in the convex hull of the roots of \( p \) over \( \mathbb{C} \)) and underlies Sturm-style root counting.

Two deeper failure modes reward attention. First, Rolle is a genuinely real theorem: for \( f(t) = e^{it} \) on \( [0, 2\pi] \) the endpoint values agree yet \( |f'| \equiv 1 \), so no complex or vector-valued Rolle can hold verbatim — what survives is only the weaker component-wise or inner-product form (e.g. for \( \mathbf{f} \colon [a,b] \to \mathbb{R}^n \) with \( \mathbf{f}(a) = \mathbf{f}(b) \), for each fixed \( \mathbf{v} \) there is \( c \) with \( \mathbf{v} \cdot \mathbf{f}'(c) = 0 \), but the point \( c \) depends on \( \mathbf{v} \)). Second, Rolle calibrates completeness: over \( \mathbb{Q} \) the theorem fails (\( (x^2-2)^2 \) above), and in reverse-mathematics style one can show that Rolle's theorem for, say, continuously differentiable functions is equivalent over a weak base theory to forms of the completeness of the reals. The theorem is not merely proved using completeness; it encodes it.

Common misconceptions. Rolle does not say the stationary point is an extremum of \( f \) on \( [a,b] \) (other stationary points may exist for other reasons); it does not locate \( c \), bound \( c \) away from the endpoints by any definite amount, or make \( c \) unique; and it imposes no differentiability requirement at \( a \) or \( b \). Finally, "differentiable on \( (a,b) \)" already implies continuity on \( (a,b) \) — the continuity hypothesis is genuinely extra only at the two endpoints, and that is exactly where the first counterexample above attacks.

Worked examples

Example 1 (direct application and location of \( c \)). Let \( f(x) = x^3 - 4x \) on \( [-2, 2] \). Show that Rolle's theorem applies and find every point it predicts.

\( f \) is a polynomial, hence continuous on \( [-2,2] \) and differentiable on \( (-2,2) \).

Polynomials are continuous and differentiable on all of \( \mathbb{R} \) (algebra of derivatives applied to \( x \mapsto x \) and constants). A

\[ f(-2) = (-8) - (-8) = 0, \qquad f(2) = 8 - 8 = 0, \qquad \text{so } f(-2) = f(2). \]

Direct evaluation; this verifies the third hypothesis. A

All hypotheses hold, so Rolle's theorem guarantees some \( c \in (-2,2) \) with \( f'(c) = 0 \).

Rolle's theorem (T-009), all three hypotheses checked in Steps 1–2. B

\[ f'(x) = 3x^2 - 4 = 0 \iff x^2 = \tfrac{4}{3} \iff x = \pm\frac{2}{\sqrt{3}} \approx \pm 1.155, \] and both values lie in \( (-2, 2) \).

Solving the derivative equation locates the guaranteed points explicitly; note Rolle promised at least one and here there are exactly two — uniqueness was never claimed. A

\[ f(x) = x^3 - 4x \text{ on } [-2,2]: \quad f'(c) = 0 \text{ at } c = -\frac{2}{\sqrt{3}} \;\text{ and }\; c = \frac{2}{\sqrt{3}}. \]

Reading. Rolle guaranteed a horizontal tangent existed before we computed anything; the computation then found both of them.

Scope. Exhaustive for this \( f \) on this interval, since \( f' \) is quadratic and has exactly these two real roots.

Example 2 (exact root count — Rolle as the "at most" half). Prove that the equation \( x^5 + 5x + 1 = 0 \) has exactly one real solution.

Let \( p(x) = x^5 + 5x + 1 \). Then \( p(-1) = -1 - 5 + 1 = -5 \lt 0 \) and \( p(0) = 1 \gt 0 \).

Direct evaluation, choosing test points with opposite signs. A

\( p \) is continuous on \( [-1, 0] \), so there exists \( r \in (-1, 0) \) with \( p(r) = 0 \): at least one real root exists.

Intermediate Value Theorem applied on \( [-1,0] \) using the sign change from Step 1. A

Suppose for contradiction \( p \) has two distinct real roots \( r_1 \lt r_2 \), so \( p(r_1) = p(r_2) = 0 \).

Setup for proof by contradiction; this creates the equal-endpoint configuration Rolle needs. B

\( p \) is a polynomial: continuous on \( [r_1, r_2] \), differentiable on \( (r_1, r_2) \), with \( p(r_1) = p(r_2) \). By Rolle's theorem there exists \( c \in (r_1, r_2) \) with \( p'(c) = 0 \).

Rolle's theorem (T-009) with \( a = r_1 \), \( b = r_2 \); every hypothesis is verified in-line. B

But \[ p'(x) = 5x^4 + 5 \ge 5 \gt 0 \quad \text{for all } x \in \mathbb{R}, \] since \( x^4 \ge 0 \). So no such \( c \) exists — contradiction. Hence \( p \) has at most one real root.

\( x^4 \ge 0 \) for real \( x \) (even power), so \( p' \) is bounded below by \( 5 \) and never vanishes; this contradicts Step 4 and refutes the assumption of Step 3. A

\[ x^5 + 5x + 1 = 0 \ \text{ has exactly one real solution, and it lies in } (-1, 0). \]

Reading. The IVT supplied "at least one"; Rolle supplied "at most one" by showing a second root would force a horizontal tangent that the everywhere-positive derivative forbids.

Scope. The template generalises: any differentiable \( f \) with \( f' \ne 0 \) throughout an interval is injective there, so sign-change plus non-vanishing derivative always yields a unique root.

Problems

(Routine.) Verify that \( f(x) = x^2 - 6x + 8 \) satisfies the hypotheses of Rolle's theorem on \( [2, 4] \), and find all values \( c \) predicted by the theorem.
Solution

\( f \) is a polynomial, hence continuous on \( [2,4] \) and differentiable on \( (2,4) \). Endpoint values: \( f(2) = 4 - 12 + 8 = 0 \) and \( f(4) = 16 - 24 + 8 = 0 \), so \( f(2) = f(4) \). All three hypotheses hold, so there exists \( c \in (2,4) \) with \( f'(c) = 0 \). Compute \( f'(x) = 2x - 6 \), so \( f'(c) = 0 \iff c = 3 \), and indeed \( 3 \in (2,4) \). Here \( c \) happens to be unique because \( f' \) is linear.
(Hypothesis diagnosis.) Let \( f(x) = 1 - x^{2/3} \) on \( [-1, 1] \). Show that \( f(-1) = f(1) \) and that there is no \( c \in (-1,1) \) with \( f'(c) = 0 \). Which hypothesis of Rolle's theorem fails, and where?
Solution

\( f(-1) = 1 - ((-1)^2)^{1/3} = 1 - 1 = 0 \) and \( f(1) = 1 - 1 = 0 \), so the endpoint values agree, and \( f \) is continuous on \( [-1,1] \) (composition of continuous functions). For \( x \ne 0 \), \[ f'(x) = -\frac{2}{3} x^{-1/3}, \] which is never zero: \( x^{-1/3} \ne 0 \) for all \( x \ne 0 \). At \( x = 0 \) the derivative does not exist: \[ \lim_{h \to 0^{+}} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^{+}} \frac{-h^{2/3}}{h} = \lim_{h \to 0^{+}} -h^{-1/3} = -\infty. \] So no \( c \in (-1,1) \) satisfies \( f'(c) = 0 \). There is no contradiction with Rolle: the hypothesis of differentiability on the open interval \( (-1,1) \) fails at the single interior point \( x = 0 \) (a cusp — which is exactly where the maximum of \( f \) sits, so the proof mechanism has nothing to grip).
(Root counting.) Prove that the equation \( x^7 + x - 1 = 0 \) has exactly one real solution.
Solution

At least one: let \( g(x) = x^7 + x - 1 \). Then \( g(0) = -1 \lt 0 \) and \( g(1) = 1 \gt 0 \); \( g \) is continuous, so by the Intermediate Value Theorem there is a root in \( (0,1) \).

At most one: suppose \( g(r_1) = g(r_2) = 0 \) with \( r_1 \lt r_2 \). \( g \) is a polynomial, so it is continuous on \( [r_1, r_2] \), differentiable on \( (r_1, r_2) \), with equal (zero) endpoint values. By Rolle's theorem some \( c \in (r_1, r_2) \) has \( g'(c) = 0 \). But \[ g'(x) = 7x^6 + 1 \ge 1 \gt 0 \quad \text{for all } x \in \mathbb{R}, \] since \( x^6 \ge 0 \). Contradiction. Hence there is exactly one real solution, and it lies in \( (0,1) \).
(Iterated Rolle.) Let \( f \colon \mathbb{R} \to \mathbb{R} \) be twice differentiable and suppose \( f(a) = f(b) = f(c) = 0 \) for some \( a \lt b \lt c \). Prove there exists \( \xi \in (a, c) \) with \( f''(\xi) = 0 \). State (without proof) the generalisation to \( n + 1 \) zeros.
Solution

Apply Rolle on \( [a, b] \): \( f \) is differentiable everywhere (hence continuous), and \( f(a) = f(b) = 0 \), so there is \( \xi_1 \in (a, b) \) with \( f'(\xi_1) = 0 \). Apply Rolle on \( [b, c] \): similarly there is \( \xi_2 \in (b, c) \) with \( f'(\xi_2) = 0 \). Now \( \xi_1 \lt b \lt \xi_2 \), and \( f' \) is differentiable on \( \mathbb{R} \) (as \( f \) is twice differentiable), hence continuous; and \( f'(\xi_1) = f'(\xi_2) = 0 \). Apply Rolle a third time, now to \( f' \) on \( [\xi_1, \xi_2] \): there exists \( \xi \in (\xi_1, \xi_2) \subseteq (a, c) \) with \( f''(\xi) = 0 \).

Generalisation: if \( f \) is \( n \) times differentiable on an interval containing \( n + 1 \) distinct zeros of \( f \), then \( f^{(n)} \) vanishes at some point strictly between the smallest and largest of those zeros. (Induction on \( n \), the step being exactly the argument above; this is the engine behind the interpolation error formula.)
(Auxiliary-function technique.) Let \( f \colon [0,1] \to \mathbb{R} \) be continuous on \( [0,1] \), differentiable on \( (0,1) \), with \( f(0) = f(1) = 0 \). Prove that there exists \( c \in (0,1) \) such that \[ f'(c) = f(c). \]
Solution

Define \( g(x) = e^{-x} f(x) \) on \( [0,1] \). Then \( g \) is continuous on \( [0,1] \) (product of continuous functions) and differentiable on \( (0,1) \) (product rule), with \[ g'(x) = e^{-x} f'(x) - e^{-x} f(x) = e^{-x}\bigl(f'(x) - f(x)\bigr). \] Endpoint values: \( g(0) = e^{0} f(0) = 0 \) and \( g(1) = e^{-1} f(1) = 0 \), so \( g(0) = g(1) \). By Rolle's theorem applied to \( g \) on \( [0,1] \), there exists \( c \in (0,1) \) with \( g'(c) = 0 \), i.e. \[ e^{-c}\bigl(f'(c) - f(c)\bigr) = 0. \] Since \( e^{-c} \gt 0 \) for every real \( c \), we may divide by it to conclude \( f'(c) - f(c) = 0 \), that is, \( f'(c) = f(c) \).

Remark. The same trick with \( g(x) = e^{-\lambda x} f(x) \) yields \( f'(c) = \lambda f(c) \) for any prescribed \( \lambda \in \mathbb{R} \): choosing the integrating factor is the whole art of Rolle-based existence proofs.