The mean value theorem

Theorem

The mean value theorem

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Statement

Let \( a, b \in \mathbb{R} \) with \( a \lt b \), and let \( f : [a,b] \to \mathbb{R} \) be a function that is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\). Then there exists at least one point \( c \in (a,b) \) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] Equivalently: the instantaneous rate of change of \(f\) equals its average rate of change over \([a,b]\) at some interior point. No continuity of \(f'\) is assumed, differentiability at the endpoints is not required, and \(f\) must be real-valued.

Why it matters

The mean value theorem (MVT) is the bridge between local information (the derivative at a point) and global information (the behaviour of \(f\) across an interval). Almost every foundational fact of single-variable calculus — that a function with zero derivative is constant, that the sign of \(f'\) controls monotonicity, that a bounded derivative forces a Lipschitz estimate, Taylor's theorem with Lagrange remainder, L'Hôpital's rule via the Cauchy form — is a corollary of this one statement. Without it, "the derivative tells you how the function grows" is a slogan; with it, it is a theorem.

It is also a model theorem for hypothesis discipline: each of its three hypotheses (continuity on the closed interval, differentiability on the open interval, real values) is exactly sharp, and dropping any one of them destroys the conclusion. Learning where the MVT breaks is as instructive as learning where it applies.

Hypotheses

\(f\) is continuous on the closed interval \([a,b]\), endpoints included.Define \( f(x) = x \) for \( 0 \le x \lt 1 \) and \( f(1) = 0 \). Then \(f\) is continuous and differentiable on \((0,1)\), and \( \frac{f(1)-f(0)}{1-0} = 0 \), yet \( f'(x) = 1 \) for every \( x \in (0,1) \). A single discontinuity at one endpoint kills the theorem, because the proof needs the extreme value theorem on all of \([a,b]\).

\(f\) is differentiable at every point of the open interval \((a,b)\).Take \( f(x) = \lvert x \rvert \) on \([-1,1]\). The average rate of change is \( \frac{1-1}{2} = 0 \), but \( f'(x) = \pm 1 \) wherever \(f'\) exists; the required point \(c\) would have to be the single non-differentiable point \(x=0\). One bad point suffices to break the conclusion.

\(f\) is real-valued (not vector- or complex-valued).Take \( f : [0, 2\pi] \to \mathbb{C} \), \( f(t) = e^{it} \). Then \( f(2\pi) - f(0) = 0 \), so the mean slope is \(0\), but \( f'(t) = i e^{it} \) has \( \lvert f'(t) \rvert = 1 \) for all \(t\) and never vanishes. The proof breaks because "maximum of \(g\)" is meaningless without the order structure of \( \mathbb{R} \); only the weaker MVT inequality \( \lVert f(b)-f(a) \rVert \le (b-a) \sup \lVert f' \rVert \) survives.

The domain is an interval \([a,b]\) of \(\mathbb{R}\) — and \(\mathbb{R}\) is complete.Over \( \mathbb{Q} \) the theorem fails: \( f : \mathbb{Q} \cap [0,2] \to \mathbb{Q} \), \( f(x) = \frac{1}{x^2 - 2} \) is continuous and differentiable at every rational point (the singularity \( \sqrt{2} \) is not in the domain) yet unbounded, so no mean value conclusion can hold. Completeness enters through the extreme value theorem, which underlies Rolle's lemma.

Proof

The proof proceeds by tilting \(f\) so that its endpoint values agree, and then applying Rolle's theorem. Because Rolle's theorem carries all the analytic content, we prove it inline (steps 4–7) from the extreme value theorem and Fermat's interior extremum theorem, rather than citing it as a black box.

\[ \text{Set } m = \frac{f(b)-f(a)}{b-a} \quad\text{and define}\quad g(x) = f(x) - f(a) - m\,(x-a) \ \text{ for } x \in [a,b]. \]

This is the chord-subtraction trick: \(g\) measures the vertical gap between \(f\) and the secant line through \((a, f(a))\) and \((b, f(b))\). Legal because \( b - a \neq 0 \), so \(m\) is a well-defined real number. B

\( g \) is continuous on \([a,b]\) and differentiable on \((a,b)\), with \( g'(x) = f'(x) - m \) on \((a,b)\).

Sums and scalar multiples of continuous functions are continuous (algebra of limits), and the affine map \( x \mapsto f(a) + m(x-a) \) is differentiable everywhere with derivative \(m\); the difference rule for derivatives gives \( g' = f' - m \). A

\[ g(a) = f(a) - f(a) - m \cdot 0 = 0, \qquad g(b) = f(b) - f(a) - \frac{f(b)-f(a)}{b-a}(b-a) = 0. \]

Direct evaluation. The whole point of the choice of \(m\) in step 1 was to force \( g(a) = g(b) \), placing \(g\) in the scope of Rolle's theorem. A

\( g \) attains a maximum value \( M \) and a minimum value \( \mu \) on \([a,b]\): there exist \( p, q \in [a,b] \) with \( g(p) = M = \sup_{[a,b]} g \) and \( g(q) = \mu = \inf_{[a,b]} g \).

This is the extreme value theorem (Weierstrass): a continuous real-valued function on a compact interval is bounded and attains its bounds. It is here — and only here — that the completeness of \( \mathbb{R} \) and the compactness of \([a,b]\) enter. C

Case 1: \( M = \mu \). Then \( g \) is constant on \([a,b]\), so \( g'(x) = 0 \) for every \( x \in (a,b) \); choose any such \( x \) as \( c \).

If the max equals the min then \( g(x) = M \) for all \(x\); the derivative of a constant function is \(0\) directly from the definition of the derivative (the difference quotient is identically \(0\)). Since \( a \lt b \), the open interval \((a,b)\) is nonempty. A

Case 2: \( M \gt \mu \). Since \( g(a) = g(b) = 0 \), at least one of the two extremes is attained at an interior point \( c \in (a,b) \).

If both the maximum and the minimum were attained only at endpoints, then \( \{M, \mu\} \subseteq \{ g(a), g(b) \} = \{0\} \), forcing \( M = \mu = 0 \), contradicting \( M \gt \mu \). So some extremum point lies in the open interval. B

At that interior extremum, \( g'(c) = 0 \). Say \( c \) is an interior maximum: for \( h \gt 0 \) small, \( \frac{g(c+h)-g(c)}{h} \le 0 \), and for \( h \lt 0 \) small, \( \frac{g(c+h)-g(c)}{h} \ge 0 \). Letting \( h \to 0^{+} \) and \( h \to 0^{-} \) gives \( g'(c) \le 0 \) and \( g'(c) \ge 0 \), hence \( g'(c) = 0 \).

Fermat's interior extremum theorem, proved in full: the one-sided difference quotients have opposite signs at an interior max, and both one-sided limits exist and equal \( g'(c) \) because \(g\) is differentiable at \(c\); order is preserved under limits. The interior-minimum case is identical with inequalities reversed. Note this needs \( c \in (a,b) \): both one-sided quotients must be available. C

\[ 0 = g'(c) = f'(c) - m \quad\Longrightarrow\quad f'(c) = m = \frac{f(b)-f(a)}{b-a}, \qquad c \in (a,b). \]

Combine step 2 with the conclusion \( g'(c) = 0 \) of steps 5–7 and solve for \( f'(c) \). This completes the proof. \( \blacksquare \) A

Result

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

Reading. If a graph is unbroken over a closed interval and has a tangent line at every interior point, then somewhere strictly between the endpoints the tangent is parallel to the chord joining the endpoint values. A car that averages 60 km/h over a journey must, at some instant, be travelling at exactly 60 km/h.

Scope. Real-valued functions of one real variable on a compact interval \([a,b]\), \( a \lt b \). Continuity is needed up to and including both endpoints; differentiability only on the interior, and \(f'\) need not be continuous or bounded. The point \(c\) is not unique in general and the theorem gives no way to locate it — it is a pure existence statement. For vector- or complex-valued functions only the mean value inequality holds.

Corollaries & converses

Zero derivative \( \Rightarrow \) constant. If \( f'(x) = 0 \) for all \( x \) in an interval \(I\), then \(f\) is constant on \(I\): for any \( x \lt y \) in \(I\), MVT gives \( f(y) - f(x) = f'(c)(y-x) = 0 \). This underwrites the "\(+C\)" of antidifferentiation. (False on a disconnected domain: a locally constant function on \( (0,1) \cup (2,3) \) with two different values has zero derivative everywhere.)
Monotonicity test. \( f' \gt 0 \) on \((a,b)\) \( \Rightarrow \) \(f\) strictly increasing on \([a,b]\); \( f' \ge 0 \) \( \Rightarrow \) nondecreasing; similarly for negative signs. (Converse fails for the strict version: \( f(x) = x^3 \) is strictly increasing but \( f'(0) = 0 \).)
Lipschitz bound. If \( \lvert f'(x) \rvert \le M \) on \((a,b)\) then \( \lvert f(x) - f(y) \rvert \le M \lvert x - y \rvert \) for all \( x, y \in [a,b] \); in particular \(f\) is uniformly continuous.
Cauchy (extended) MVT. For \( f, g \) continuous on \([a,b]\), differentiable on \((a,b)\): \( \exists\, c \) with \( \left( f(b) - f(a) \right) g'(c) = \left( g(b) - g(a) \right) f'(c) \) — the engine behind L'Hôpital's rule and the Lagrange remainder in Taylor's theorem.
Converse fails. The MVT does not run backwards: not every interior point is a mean value point for some interval. For \( f(x) = x^3 \), the point \( c = 0 \) satisfies \( f'(0) = 0 \), but \( \frac{b^3 - a^3}{b - a} = a^2 + ab + b^2 = 0 \) has no solution with \( a \lt b \) real other than degenerate \( a = b = 0 \); no genuine chord of \( x^3 \) is horizontal. Existence of \(c\) given \([a,b]\) is a theorem; existence of \([a,b]\) given \(c\) is false.

Fails without

Drop endpoint continuity: \( f(x) = x \) on \( [0,1) \), \( f(1) = 0 \). Mean slope \( = 0 \), but \( f' \equiv 1 \) on \((0,1)\). No point \(c\) exists. The extreme value theorem fails for this \(g\), and with it the whole proof.
Drop differentiability at one interior point: \( f(x) = \lvert x \rvert \) on \([-1,1]\). Mean slope \( = 0 \), but \( f'(x) \in \{-1, +1\} \) wherever defined. The unique candidate for \(c\) is the corner \( x = 0 \), precisely where \(f'\) does not exist.
Drop real-valuedness: \( f(t) = e^{it} = (\cos t, \sin t) \) on \([0, 2\pi]\). Then \( f(2\pi) = f(0) \) but \( \lvert f'(t) \rvert = 1 \) never vanishes: a particle can return to its start without ever having velocity zero, because velocity is a vector and its components vanish at different times.
Drop completeness of the field: on \( \mathbb{Q} \cap [0,2] \) the function \( f(x) = \frac{1}{x^2-2} \) is differentiable everywhere in its domain yet unbounded; the MVT (and every corollary, e.g. bounded derivative \( \Rightarrow \) Lipschitz) fails over \( \mathbb{Q} \).

Common errors

The unique-\(c\) fallacy. Writing "the point \(c\)" and manipulating it as if determined by \(a\) and \(b\). MVT asserts existence, never uniqueness: \( f(x) = \sin x \) on \([0, 2\pi]\) has mean slope \(0\) attained at both \( c = \pi/2 \) and \( c = 3\pi/2 \).
Demanding differentiability at the endpoints. \( f(x) = \sqrt{1 - x^2} \) on \([-1,1]\) has vertical tangents at \( \pm 1 \) and is not differentiable there — MVT still applies, since only interior differentiability is hypothesised.
Assuming \(f'\) is continuous. The theorem never says so, and \( f(x) = x^2 \sin(1/x) \) (with \( f(0)=0 \)) is differentiable everywhere with \(f'\) discontinuous at \(0\); MVT applies to it without complaint. Conversely, arguments that apply the intermediate value theorem to \(f'\) need Darboux's theorem, not continuity of \(f'\).
Confusing MVT with IVT. The intermediate value theorem is about values of \(f\); the mean value theorem is about values of \(f'\). "There is a \(c\) with \( f(c) \) equal to the average of \( f(a) \) and \( f(b) \)" is a different (IVT-flavoured) statement.
Componentwise MVT for vector functions. Applying MVT to each component of \( f : [a,b] \to \mathbb{R}^n \) yields a different \( c_i \) for each component; there is no single \(c\) working for all components simultaneously (see \( e^{it} \) above).
Using MVT across a singularity. Claiming a mean value point for \( f(x) = 1/x \) on \([-1, 1]\): the hypothesis "continuous on \([a,b]\)" fails at \(0\), so the theorem is silent — no conclusion, not a false one.

Discussion

Historically the result crystallised late. Rolle proved his root-separation lemma for polynomials in 1691 — ironically, as part of a critique of the infinitesimal calculus he distrusted. Lagrange stated the mean value theorem in essentially modern form in 1797, using it to found his (ultimately unsuccessful) programme of basing calculus on power series; Cauchy made it the workhorse of his 1823 Calcul infinitésimal, and the modern proof via an auxiliary function and Rolle is usually credited to Ossian Bonnet in the mid-nineteenth century. The clean statement thus postdates by more than a century the calculus it now underpins.

Structurally, the MVT sits at the junction of three ideas: compactness (the extreme value theorem, step 4), the order structure of \( \mathbb{R} \) (Fermat's sign argument, step 7), and an affine normalisation (the tilt in step 1). This explains its failure modes precisely: complex- and vector-valued functions lose the order structure, \( \mathbb{Q} \) loses compactness/completeness, and disconnected domains lose the single chord to tilt against. Every hypothesis is traceable to a specific step of the proof — a useful habit of reading to cultivate.

Many modern analysts, following Dieudonné, argue that the equality form of the MVT is a stepping stone and the mean value inequality \( \lvert f(b) - f(a) \rvert \le (b-a) \sup_{(a,b)} \lvert f' \rvert \) is the result one actually uses: it is what survives in Banach spaces, what proves uniqueness in Picard–Lindelöf, and what drives error estimates in numerical analysis. Nearly every application in this page's Worked examples and Problems secretly uses only the inequality. The equality form remains indispensable, however, for exact identities: Taylor's theorem with Lagrange remainder \( f(b) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(b-a)^k + \frac{f^{(n+1)}(\xi)}{(n+1)!}(b-a)^{n+1} \) is the MVT iterated, and reduces to it when \( n = 0 \).

Two refinements are worth knowing at the rigorous level. First, Darboux's theorem: a derivative on an interval satisfies the intermediate value property even when discontinuous — so \(f'\) can never have jump discontinuities, only essential ones (like \( f'(x) = 2x\sin(1/x) - \cos(1/x) \) at \(0\)). Second, the hypothesis "differentiable on \((a,b)\)" can be weakened: the conclusion survives if \(f\) is continuous on \([a,b]\) and differentiable off a countable set, provided one settles for the mean value inequality with \( \sup \) replaced by an essential bound — a first step toward the absolutely continuous functions of measure theory, where \( f(b) - f(a) = \int_a^b f' \) restores an exact mean value in integral form. Common misconceptions. The MVT is not "obvious from the picture": the picture assumes the graph has a tangent everywhere and that a tilted maximum exists, which are exactly the analytic hypotheses. And the theorem does not say \(c\) is computable or unique — most applications never locate \(c\) and only use bounds on \( f' \) over the whole interval.

Worked examples

Example 1. The sine function is 1-Lipschitz: \( \lvert \sin x - \sin y \rvert \le \lvert x - y \rvert \) for all \( x, y \in \mathbb{R} \), with equality only when \( x = y \).

Fix \( x \neq y \); without loss of generality \( y \lt x \). Apply the MVT to \( f(t) = \sin t \) on \([y, x]\).

\( \sin \) is continuous and differentiable on all of \( \mathbb{R} \), hence on \([y,x]\) and \((y,x)\); both hypotheses hold. A

\[ \exists\, c \in (y, x): \quad \frac{\sin x - \sin y}{x - y} = \cos c. \]

Direct application of the MVT with \( f' = \cos \). A

\[ \lvert \sin x - \sin y \rvert = \lvert \cos c \rvert \cdot \lvert x - y \rvert \le 1 \cdot \lvert x - y \rvert. \]

\( \lvert \cos c \rvert \le 1 \) for every real \(c\). For strictness when \( x \neq y \): equality would force \( \lvert \cos c \rvert = 1 \) at the specific mean value point, but if \( \lvert \sin x - \sin y \rvert = \lvert x - y \rvert \gt 0 \) one checks (e.g. on \( \lvert x - y \rvert \lt \pi \), splitting longer intervals) that \( \cos \) cannot maintain \( \lvert \cos t \rvert = 1 \) on a nondegenerate interval, whereas the sharper integral form \( \sin x - \sin y = \int_y^x \cos t \, dt \) makes strictness immediate. B

\[ \lvert \sin x - \sin y \rvert \le \lvert x - y \rvert \quad \forall\, x, y \in \mathbb{R} \]

Reading. A bound on the derivative (\( \lvert \cos \rvert \le 1 \)) transfers, via the MVT, to a global Lipschitz bound on the function — no explicit knowledge of the point \(c\) was needed, only an estimate valid at every candidate \(c\).

Scope. The same one-line argument shows any \(f\) with \( \lvert f' \rvert \le M \) on an interval satisfies \( \lvert f(x)-f(y) \rvert \le M\lvert x-y \rvert \) there.

Example 2. Estimating \( \sqrt{66} \) without a calculator: show \( 8 + \frac{1}{9} \lt \sqrt{66} \lt 8 + \frac{1}{8} \).

Apply the MVT to \( f(x) = \sqrt{x} \) on \([64, 66]\).

\( f \) is continuous on \([64,66]\) and differentiable on \((64,66)\) with \( f'(x) = \frac{1}{2\sqrt{x}} \), since \( x \gt 0 \) throughout. Hypotheses verified. A

\[ \exists\, c \in (64, 66): \quad \sqrt{66} - \sqrt{64} = \frac{66 - 64}{2\sqrt{c}} = \frac{1}{\sqrt{c}}, \quad\text{i.e.}\quad \sqrt{66} = 8 + \frac{1}{\sqrt{c}}. \]

MVT with \( f(66) - f(64) = f'(c)(66-64) \) and \( \sqrt{64} = 8 \). A

\[ 64 \lt c \lt 66 \lt 81 \ \Longrightarrow\ 8 \lt \sqrt{c} \lt 9 \ \Longrightarrow\ \frac{1}{9} \lt \frac{1}{\sqrt{c}} \lt \frac{1}{8}. \]

\( \sqrt{\cdot} \) is strictly increasing (its derivative is positive — itself a corollary of the MVT), and \( t \mapsto 1/t \) is strictly decreasing on \( (0, \infty) \). We bounded \( \sqrt{c} \) using the perfect squares bracketing \(c\), so no circularity: we never needed the value of \( \sqrt{66} \) itself. B

\[ 8.111\overline{1} = 8 + \tfrac{1}{9} \ \lt\ \sqrt{66} \ \lt\ 8 + \tfrac{1}{8} = 8.125 \]

Reading. The MVT converts "\(\sqrt{66}\) is near \(\sqrt{64}\)" into a rigorous two-sided error bound: the increment equals derivative-at-\(c\) times step, and bounding the derivative over the whole interval brackets the answer. (True value: \( \sqrt{66} \approx 8.1240 \).)

Scope. This is the prototype of every first-order error estimate in numerical analysis; Taylor's theorem with Lagrange remainder refines it to higher order.

Problems

Verify the mean value theorem explicitly for \( f(x) = x^2 \) on \([1,3]\): check the hypotheses, and find all points \(c\) that satisfy the conclusion.
Solution
\( f(x) = x^2 \) is a polynomial, hence continuous on \([1,3]\) and differentiable on \((1,3)\); both hypotheses hold. The mean slope is \( \frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4 \). The conclusion requires \( f'(c) = 2c = 4 \), so \( c = 2 \), which indeed lies in the open interval \((1,3)\). Here \(c\) happens to be unique because \( f' \) is strictly monotone (injective); for general \(f\) uniqueness can fail.
Prove that for all \( x \gt 0 \), \[ \frac{x}{1+x} \lt \ln(1+x) \lt x. \]
Solution
Fix \( x \gt 0 \) and apply the MVT to \( f(t) = \ln(1+t) \) on \([0, x]\): \(f\) is continuous there and differentiable on \((0,x)\) with \( f'(t) = \frac{1}{1+t} \). Hence there is \( c \in (0, x) \) with \[ \ln(1+x) = \ln(1+x) - \ln 1 = \frac{x}{1+c}. \] Since \( 0 \lt c \lt x \), we have \( 1 \lt 1 + c \lt 1 + x \), so \( \frac{1}{1+x} \lt \frac{1}{1+c} \lt 1 \). Multiplying by \( x \gt 0 \) preserves the strict inequalities: \[ \frac{x}{1+x} \lt \frac{x}{1+c} = \ln(1+x) \lt x. \qquad \blacksquare \]
Let \( f : (a,b) \to \mathbb{R} \) be differentiable with \( \lvert f'(x) \rvert \le M \) for all \( x \in (a,b) \). Prove that \(f\) is uniformly continuous on \((a,b)\), and deduce that \(f\) extends continuously to \([a,b]\).
Solution
Lipschitz bound. For any \( x \lt y \) in \((a,b)\), \(f\) is continuous on \([x,y]\) and differentiable on \((x,y)\) (restriction of the hypothesis), so the MVT gives \( c \in (x,y) \) with \( f(y)-f(x) = f'(c)(y-x) \), whence \( \lvert f(y)-f(x) \rvert \le M \lvert y - x \rvert \). Uniform continuity. Given \( \varepsilon \gt 0 \), take \( \delta = \varepsilon / (M+1) \); then \( \lvert x - y \rvert \lt \delta \) implies \( \lvert f(x)-f(y) \rvert \le M\delta \lt \varepsilon \). (Using \(M+1\) covers the case \(M = 0\).) Extension. Let \( (x_n) \subseteq (a,b) \) with \( x_n \to b \). Then \( (x_n) \) is Cauchy, and the Lipschitz bound gives \( \lvert f(x_n) - f(x_m) \rvert \le M \lvert x_n - x_m \rvert \), so \( (f(x_n)) \) is Cauchy in \( \mathbb{R} \) and converges by completeness. If \( (y_n) \) is another sequence tending to \(b\), then \( \lvert f(x_n) - f(y_n) \rvert \le M\lvert x_n - y_n \rvert \to 0 \), so the limit is independent of the sequence; define \( f(b) \) to be this common limit, and similarly at \(a\). Sequential characterisation of limits shows the extended \(f\) is continuous on \([a,b]\). \( \blacksquare \)
Suppose \( f : \mathbb{R} \to \mathbb{R} \) is differentiable and \( f'(x) \to L \) as \( x \to \infty \). Prove that \( f(x+1) - f(x) \to L \) as \( x \to \infty \). Show by example that the converse fails.
Solution
For each \( x \), \(f\) is continuous on \([x, x+1]\) and differentiable on \((x, x+1)\), so the MVT provides \( \xi_x \in (x, x+1) \) with \[ f(x+1) - f(x) = f'(\xi_x)\,\left( (x+1) - x \right) = f'(\xi_x). \] Given \( \varepsilon \gt 0 \), choose \( R \) such that \( \lvert f'(t) - L \rvert \lt \varepsilon \) for all \( t \gt R \). For \( x \gt R \) we have \( \xi_x \gt x \gt R \), hence \( \lvert f(x+1)-f(x) - L \rvert = \lvert f'(\xi_x) - L \rvert \lt \varepsilon \). Thus \( f(x+1)-f(x) \to L \). Converse fails: let \( f(x) = \sin(2\pi x) \). Then \( f(x+1) - f(x) = 0 \to 0 \), but \( f'(x) = 2\pi \cos(2\pi x) \) oscillates and has no limit. So the unit increments can converge while \( f' \) does not. \( \blacksquare \)
(Harder.) Let \( f : [0,1] \to \mathbb{R} \) be continuous on \([0,1]\), differentiable on \((0,1)\), with \( f(0) = 0 \) and \( f(1) = 1 \). Prove there exist distinct points \( c_1, c_2 \in (0,1) \) such that \[ \frac{1}{f'(c_1)} + \frac{1}{f'(c_2)} = 2. \]
Solution
Since \(f\) is continuous with \( f(0) = 0 \lt \frac{1}{2} \lt 1 = f(1) \), the intermediate value theorem yields \( t \in (0,1) \) with \( f(t) = \frac{1}{2} \). Apply the MVT on \([0, t]\): there is \( c_1 \in (0, t) \) with \[ f'(c_1) = \frac{f(t) - f(0)}{t - 0} = \frac{1/2}{t} = \frac{1}{2t}, \] and on \([t, 1]\): there is \( c_2 \in (t, 1) \) with \[ f'(c_2) = \frac{f(1) - f(t)}{1 - t} = \frac{1/2}{1-t} = \frac{1}{2(1-t)}. \] Both values are the exhibited positive reals, so in particular \( f'(c_1) \neq 0 \neq f'(c_2) \) and the reciprocals are defined: \[ \frac{1}{f'(c_1)} + \frac{1}{f'(c_2)} = 2t + 2(1-t) = 2. \] Finally \( c_1 \lt t \lt c_2 \), so \( c_1 \neq c_2 \). \( \blacksquare \) (The same device with IVT points \( f(t_k) = k/n \) produces \(n\) distinct points with \( \sum_{k=1}^{n} \frac{1}{f'(c_k)} = n \).)