Rolle's Theorem to MVT: A Guided Walk Through a 12-Problem Calculus Worksheet

You have a classwork sheet in front of you — classwork9.pdf, two pages, ten concept bullets on page 1 and twelve numbered problems on page 2. The sheet states Rolle's Theorem, the Mean Value Theorem, the First and Second Derivative Tests, a concavity criterion, an inflection-point definition, and a seven-step curve-sketching procedure, then asks you to use these tools on twelve specific functions. Every problem is unlocked by a theorem the sheet states without proof. The job of this tutorial is to provide those proofs and to walk each problem end-to-end, in the order the worksheet asks it.

Two pieces of vocabulary you'll meet over and over: a critical point is an interior point c where f'(c) = 0 or f'(c) does not exist, and the secant line is the straight line through two points (a, f(a)) and (b, f(b)) on the curve, with slope (f(b) − f(a)) / (b − a). The MVT, in particular, says: somewhere between a and b, the curve has a tangent whose slope equals that secant slope.

The sheet itself never proves any of its ten bullets, but it expects you to recognise which hypothesis fails when the conclusion fails, which corollary turns a local derivative sign into a global conclusion, and which row of its own 3×3 derivative-meaning grid (page 2, figure 1) corresponds to the case at hand. We will revisit that grid several times; it is the at-a-glance glossary for the entire second half of the worksheet.classwork

1. What the worksheet assumes you can already do, and what it is about to ask you to prove

Two theorems start everything. Rolle's Theorem is the restrictive case where the two endpoints have the same height, forcing at least one horizontal tangent in between. The Mean Value Theorem (MVT) drops the equal-height requirement and lets the secant line tilt; it then guarantees a tangent somewhere that is parallel to that slanted secant.

We open with the worksheet's own MVT figure, because it shows the geometric picture you will read off every later problem.

Figure 1 (from the worksheet, page 1, §2): the blue curve is y = f(x). The red line is the tangent at the interior point c, labelled Slope f'(c). The yellow line is the secant through A = (a, f(a)) and B = (b, f(b)), labelled Slope [f(b) − f(a)] / (b − a). The dashed vertical line marks x = c. The figure shows the geometric content of MVT in one panel: the two lines are parallel.classwork

The two formal statements the sheet uses are these. Rolle's Theorem: if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists some c ∈ (a, b) with f'(c) = 0. Mean Value Theorem: if f is continuous on [a, b] and differentiable on (a, b), then there exists some c ∈ (a, b) with f'(c) = (f(b) − f(a)) / (b − a).classwork

Notice the asymmetric hypotheses: continuity is required on the closed interval [a, b], but differentiability is only required on the open interval (a, b). That split is not accidental, and it is exactly what Problem 1 will ask you to test. It is also what makes Rolle's Theorem a special case of MVT: setting f(a) = f(b) in MVT gives the secant slope 0, so the parallel tangent must have slope 0, and f'(c) = 0 falls out.openstax

OpenStax's Example 4.15 — verify MVT for f(x) = √x on [0, 9] — gives a concrete instance where MVT actually produces a numerical c. The secant slope is (3 − 0)/(9 − 0) = 1/3, so we need f'(c) = 1/(2√c) = 1/3, giving c = 9/4. We will use this same function in Chapter 3 as the worked example for the auxiliary-function proof.openstax

Now here is the key visual point: the worksheet's figure shows one specific c, but MVT only guarantees at least one such c. On some curves there can be two, three, or infinitely many. The LibreTexts adaptation of Stewart's Figure 3 captures this with a wavy curve and two interior points c₁, c₂ at which the tangent line matches the secant slope.stewart

So the worksheet expects you to leave this chapter knowing three things: the formal statements of Rolle and MVT, the geometric picture (tangent at c parallel to secant through the endpoints), and the fact that Rolle's Theorem is MVT in the special case f(a) = f(b). Problem 1 asks you to find counterexamples that prove each hypothesis is necessary. Problem 3 asks you to prove the parallel-tangents theorem using Rolle directly. We open those in the next two chapters.

2. Why does Rolle's conclusion sometimes fail — and what does each hypothesis actually buy you?

The cleanest proof of Rolle's Theorem is the one OpenStax gives: by the Extreme Value Theorem (EVT), a continuous function on a closed bounded interval attains both a maximum and a minimum somewhere on that interval; if f is constant on (a, b) then f'(c) = 0 everywhere and we are done; otherwise either the maximum or the minimum occurs at an interior point c ∈ (a, b), and at that interior extremum Fermat's Theorem (differentiable local extremum ⇒ derivative is zero) gives f'(c) = 0.openstax

Now suppose we drop one hypothesis. The conclusion can break. Problem 1 of the worksheet asks you to produce a function with f(a) = f(b) (or, in part (c), with no constraint on the endpoints) and no c ∈ (a, b) with f'(c) = 0. The three parts correspond to dropping the continuity hypothesis, the interior-differentiability hypothesis, and the equal-endpoint hypothesis, respectively. The CalcLearn article "When Rolle's Theorem Fails" maps one counterexample per dropped hypothesis, exactly as Problem 1's three parts ask:calclearn

The middle row is the visual anchor for the whole chapter. OpenStax's Figure 4.22 plots the same V-shaped counterexample, with the explicit annotation "No c such that f'(c) = 0":

Figure 2 (OpenStax Figure 4.22): the blue V-shape is f(x) = |x| − 1, with vertex at (0, −1) and the curve passing through (−1, 0) and (1, 0) on the x-axis. The function is continuous on [−1, 1] and the endpoint values match, but the corner at 0 is not differentiable. The annotation reads "No c such that f'(c) = 0": the conclusion of Rolle's Theorem fails because one of its three hypotheses (differentiability on (a, b)) is missing.openstax

The positive case — Rolle does give a horizontal tangent — is worth seeing too. The LibreTexts adaptation of Stewart Figure 1 stacks three panels of functions that satisfy all three hypotheses:

Figure 3 (LibreTexts adaptation of Stewart Figure 1, three panels): each panel shows a function on [a, b] whose endpoints lie on a common horizontal dashed reference line (so f(a) = f(b)), and each panel marks an interior point c (or two) where the tangent is horizontal. Panel (a): a downward arc with a single horizontal tangent at the peak. Panel (b): an upward arc with a single horizontal tangent at the trough. Panel (c): one period of a sine-like wave, with two interior horizontal tangents, one at the peak and one at the trough. The figure illustrates two facts at once: the conclusion f'(c) = 0 holds for every function that satisfies all three hypotheses, and the c is not necessarily unique.stewart

One last subtlety: the asymmetry in Rolle's hypotheses (continuous on closed [a, b], differentiable on open (a, b)) is the least restrictive form the theorem can take. The half-circle f(x) = √(r² − x²) on [-r, r] has vertical tangents at the endpoints — f is not differentiable at ±r — but Rolle's Theorem still applies because Rolle only needs differentiability on the open interval.wikipedia The square-root function √x on [0, 1] has the same property. Dropping the endpoint-differentiability requirement is exactly what makes Rolle general enough to apply to the parallel-tangents problem on the next page.

3. How do you prove MVT, and how does that proof unlock Problems 3 and 4?

The standard proof of MVT is a one-step reduction to Rolle. Define an auxiliary function:

$$g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a},(x - a)$$

Geometrically, g(x) is the vertical distance from the curve y = f(x) to the secant line through (a, f(a)) and (b, f(b)). The constants are chosen so that g(a) = g(b) = 0. The function g inherits continuity on [a, b] and differentiability on (a, b) from f, so Rolle's Theorem gives some c ∈ (a, b) with g'(c) = 0. Now g'(x) = f'(x) - (f(b) − f(a)) / (b − a), so g'(c) = 0 rearranges to f'(c) = (f(b) − f(a)) / (b − a), which is exactly MVT.openstax

This proof gives more than MVT itself — it gives the three MVT corollaries that Problems 4 and 8 (and Problem 2's uniqueness argument) depend on.openstax

Corollary 1 (constant-from-zero-derivative). If f'(x) = 0 for every x ∈ (a, b), then f is constant on (a, b). The proof is by contradiction: pick any two points x₁ < x₂ in (a, b); if f(x₁) ≠ f(x₂), MVT on [x₁, x₂] gives a c ∈ (x₁, x₂) with f'(c) = (f(x₂) − f(x₁)) / (x₂ − x₁) ≠ 0, contradicting f' ≡ 0. This is exactly the hint for Problem 4(a): Apply MVT to [x₁, x₂] ⊆ (a, b).

Corollary 2 (differ-by-constant). If f'(x) = g'(x) for every x ∈ (a, b), then f(x) = g(x) + C for some constant C. Apply Corollary 1 to h = f − g: since h'(x) = f'(x) − g'(x) = 0 everywhere on (a, b), h is constant, so f = g + C. This solves Problem 4(b).

Corollary 3 (increasing-from-positive-derivative). If f'(x) > 0 for every x ∈ (a, b), then f is increasing on [a, b]. Proof by contradiction: assume f(a) ≥ f(b); MVT on [a, b] gives c ∈ (a, b) with f'(c) = (f(b) − f(a)) / (b − a) ≤ 0, contradicting f' > 0. The corresponding statement with f'(x) < 0 gives f strictly decreasing.

A worked example makes the theorem concrete. OpenStax's Example 4.15 — verify MVT for f(x) = √x on [0, 9] — is the worked numerical instance we will use here; OpenStax also gives Example 4.16 (rock dropped, t = 5/4) as a second illustration of MVT producing an explicit c:

Figure 4 (OpenStax Figure 4.27): the blue curve is f(x) = √x from (0, 0) up to (9, 3). The black dot at (9/4, 3/2) is the unique interior point where MVT's c lies. The two orange dashed segments through that dot are the tangent line; both endpoints of the secant, (0, 0) and (9, 3), are marked. Secant slope 1/3 and tangent slope f'(9/4) = 1/(2 · 3/2) = 1/3 agree, as the theorem guarantees.openstax

Problem 3 of the worksheet asks the parallel-tangents theorem: given f, g continuous on (a, b) and differentiable on [a, b] with f(a) = g(a) and f(b) = g(b), prove there is a c ∈ (a, b) with f'(c) = g'(c). A natural approach uses the auxiliary-function trick, the same Rolle-to-MVT machinery from the proof above: define h = f − g. Then h(a) = h(b) = 0, and h is continuous on [a, b] and differentiable on (a, b), so Rolle's Theorem applies to h, giving some c ∈ (a, b) with h'(c) = 0, that is, f'(c) = g'(c). At that point c, the tangent lines of f and g have the same slope — they are parallel.classwork

The deeper lesson is that Rolle's Theorem and MVT are not separate results you have to memorise independently — they are the same result at different orientations, and the auxiliary-function trick is the bridge between them. We will use that bridge repeatedly in the next chapter.

4. What can MVT tell you about f(5) when all you know is f(1) = 10 and f' ∈ [−1, 2]?

Problem 6 of the worksheet is the cleanest application of MVT to a numerical answer. We are given f differentiable on [1, 5], f(1) = 10, and f'(x) ∈ [−1, 2] for every x in some neighbourhood. We want bounds on f(5).

By MVT on [1, 5], there exists c ∈ (1, 5) with f(5) − f(1) = f'(c) · (5 − 1) = 4 f'(c). Since f'(c) ∈ [−1, 2], we get 4 f'(c) ∈ [−4, 8], hence f(5) = 10 + 4 f'(c) ∈ [6, 18].openstax

Both bounds are sharp: if f'(x) ≡ −1 everywhere on [1, 5], then f(5) = 10 − 4 = 6 (this is f(x) = −x + 11); if f'(x) ≡ 2 everywhere, then f(5) = 10 + 8 = 18 (this is f(x) = 2x + 8). So [6, 18] is not just a crude bracket — no tighter bounds are possible without more information.stewart

This generalises. The bounding corollary in OpenStax's §4.4 reads:

If |f'(x)| ≤ M for all x ∈ [a, b], then |f(b) − f(a)| ≤ M · |b − a|.openstax

The arithmetic template is identical to Stewart's §4.2 Example 5 (different constants: f(0) = −3, f' ≤ 5, find the largest possible f(2), giving 7).stewart

The same machinery, with a different sign, unlocks Problem 2: show x³ + x + 1 = 0 has exactly one real root. Let f(x) = x³ + x + 1. Then f'(x) = 3x² + 1 > 0 everywhere on ℝ. By MVT Corollary 3, f is strictly increasing on ℝ, so any horizontal line meets the graph at most once. In particular y = 0 meets the graph at most once, and since f(−1) = −1 < 0 and f(0) = 1 > 0, by the Intermediate Value Theorem there is exactly one real root. This is a complete proof; no auxiliary function needed.

The same corollary unlocks Problem 8: find the monotonic intervals of f(x) = x³ − 12x − 5. Compute f'(x) = 3x² − 12 = 3(x − 2)(x + 2). The sign chart of f'(x) is:

• f'(x) > 0 on (−∞, −2) and (2, ∞) → f is increasing on these intervals.
• f'(x) < 0 on (−2, 2) → f is decreasing on this interval.

Each conclusion is an immediate application of Corollary 3 (and its f' < 0 mirror). The worksheet's §3 remark is worth flagging here: the monotonicity corollary extends to infinite intervals like (−∞, −2) and (2, ∞), not just closed bounded ones, because the proof only uses the sign of f' throughout the interior — it does not need finite endpoints.classwork

So MVT is the bridge from local derivative information to global conclusions about a function's behaviour. Three corollaries: f' ≡ 0 ⇒ constant (Problem 4a); f' = g' ⇒ f − g constant (Problem 4b); f' > 0 ⇒ strictly increasing (Problems 2, 8). And one sharp numerical bound: f(5) ∈ [6, 18] when f(1) = 10 and f' ∈ [−1, 2] (Problem 6). The next chapter takes the sign of f' and turns it into extrema.

5. How does the sign of f' classify a critical point — and when is the First Derivative Test inconclusive?

The First Derivative Test reads the sign of f' on either side of a critical point c:

• f' changes from negative to positive across c ⇒ c is a local minimum.
• f' changes from positive to negative across c ⇒ c is a local maximum.
• f' keeps the same sign on both sides ⇒ c is not a local extremum.classwork

This is the sheet's own 3×3 grid, row 3 left cell:

Figure 5 (the worksheet's own 3×3 derivative-meaning grid): row 1 covers the meaning of y' (differentiability, rising, falling). Row 2 covers concavity and inflection points from y''. Row 3 covers the First Derivative Test (cell 7: y' changes sign ⇒ local max or local min) and the Second Derivative Test (cells 8 and 9: y'=0 and y''<0 ⇒ local max; y'=0 and y''>0 ⇒ local min). The grid is the at-a-glance glossary the sheet wants you to internalise before starting Problems 8–12.classwork

For Problem 9(a) — find the absolute extrema of f(x) = x^{4/3} − 4x^{1/3} on ℝ — the First Derivative Test has to be applied at a non-smooth critical point, which is the subtle case the worksheet is testing. Differentiate:

$$f'(x) = \tfrac{4}{3} x^{1/3} - \tfrac{4}{3} x^{-2/3} = \tfrac{4}{3} \cdot \frac{x - 1}{x^{2/3}}$$

There are two critical points: x = 1 (where f'(1) = 0) and x = 0 (where f'(0) does not exist, because of the x^{-2/3} factor). Evaluate f at both: f(1) = 1^{4/3} − 4 · 1^{1/3} = 1 − 4 = −3; f(0) = 0 − 0 = 0. As x → ±∞, x^{4/3} dominates −4 x^{1/3}, so f(x) → +∞ at both ends. By the First Derivative Test on either side of x = 1: on (0, 1), f'(x) < 0 (numerator negative, denominator positive), so f is decreasing; on (1, ∞), f'(x) > 0, so f is increasing — x = 1 is a local minimum, and since the function goes to +∞ at both ends, this is the absolute minimum (value −3).openstax

The critical point x = 0 has f'(0) undefined — f has a vertical tangent there, not a regular derivative. The First Derivative Test still works at such a cusp point: read the sign of f' on either side of 0. For x < 0, f'(x) < 0 (numerator x − 1 < 0, denominator x^{2/3} > 0); for x > 0 small, numerator x − 1 < 0 and denominator x^{2/3} > 0, so f'(x) < 0 too. Same sign on both sides ⇒ x = 0 is not a local extremum. (OpenStax's Example 4.18, 5x^{1/3} − x^{5/3}, is the canonical worked example for this case at an undefined-derivative critical point: f'(0) is undefined, but f' has the same sign (positive) on both sides of x = 0, so the First Derivative Test classifies it as not an extremum.openstax

The wider lesson: a critical point is any interior point where f'(c) = 0 or f'(c) does not exist. The First Derivative Test does not care which kind — it only reads the sign of f' on either side, so it handles both smooth and non-smooth critical points in one rule. The Second Derivative Test, by contrast, only reads f''(c), and as we will see in the next chapter, it can fail even when a smooth extremum exists.

6. What does the sign of f'' tell you, and why is x⁴ at 0 the case that breaks the Second Derivative Test?

Concavity has two equivalent definitions. The sheet's §5 says f is concave up on an open interval I if f' is increasing on I; concave down if f' is decreasing on I. The sheet's §6 then says: if f'' > 0 on I, then f is concave up; if f'' < 0, then f is concave down.classwork

The two definitions are equivalent because of MVT applied to f': f'' > 0 on I means f' is differentiable with positive derivative on I, which by MVT Corollary 3 (with f' in place of f) means f' is increasing on I. That's the sheet's §5 definition of concave up. Conversely, if f' is increasing and differentiable, f'' ≥ 0 everywhere on I.openstax

An inflection point is a point c where f'(c) exists and the concavity actually changes at c (sheet §7). The sheet's remark is critical: f''(c) = 0 or f''(c) does not exist is necessary for an inflection point, but not sufficient.classwork

The canonical counterexample is f(x) = x⁴ at x = 0. Compute f''(0) = 12 · 0² = 0, so the necessary condition is met — but the concavity does not change there. f(x) = x⁴ is locally concave up on both sides of 0 (you can check f''(x) = 12x² > 0 for x ≠ 0, and the concavity test in OpenStax's Theorem 4.10 (Concavity Test) says this f'' > 0 makes f concave up everywhere except possibly at 0). So x⁴ at 0 is not an inflection point even though f''(0) = 0.openstax

Problem 10 walks four examples to make this concrete.

(a) f(x) = cos x. f'(x) = −sin x, f''(x) = −cos x. f''(x) > 0 when cos x < 0, i.e., on ((π/2) + 2kπ, (3π/2) + 2kπ) for integer k — these are the concave-up arcs. f''(x) < 0 when cos x > 0, i.e., on (−(π/2) + 2kπ, (π/2) + 2kπ) — these are the concave-down arcs. Concavity changes at every x = (π/2) + kπ (i.e., at odd multiples of π/2), and f is continuous at each such point, so every odd multiple of π/2 is an inflection point.

(b) f(x) = x^{5/3}. f'(x) = (5/3) x^{2/3}, f''(x) = (10/9) x^{−1/3}. f''(0) does not exist. For x < 0, x^{−1/3} < 0, so f'' < 0 (concave down); for x > 0, f'' > 0 (concave up). Concavity changes at 0, and f is continuous there, so x = 0 is an inflection point. Note the necessary condition is "f''(0) does not exist," which is one of the two cases the sheet flags.

(c) f(x) = x⁴. As discussed above: f''(x) = 12x² ≥ 0 everywhere, with f''(0) = 0. Concavity is up on both sides of 0; concavity does not change. So x = 0 is not an inflection point.

(d) f(x) = x^{1/3}. f'(x) = (1/3) x^{−2/3}, f''(x) = (−2/9) x^{−5/3}. f''(0) does not exist. For x < 0, x^{−5/3} < 0, so f'' > 0 (concave up); for x > 0, f'' < 0 (concave down). Concavity changes at 0, and f is continuous there, so x = 0 is an inflection point. This is the standard case where the necessary condition "f''(c) does not exist" is met and the conclusion actually follows.

The Second Derivative Test for extrema is a different story. It states: if f'(c) = 0 and f''(c) < 0, then c is a local maximum; if f'(c) = 0 and f''(c) > 0, then c is a local minimum. But the test is inconclusive when f''(c) = 0. Three canonical examples all have f''(0) = 0 and differ wildly: x³ at 0 has no extremum (the function is monotonically increasing); x⁴ at 0 has a local minimum; −x⁴ at 0 has a local maximum. The Second Derivative Test gives no verdict here — you have to fall back on the First Derivative Test and read the sign of f'.openstax

This is exactly the case the worksheet's 3×3 grid (Figure 5 above) leaves implicit: row 3 cells 8 and 9 only cover the cases f'' < 0 and f'' > 0. The third case — f'' = 0 at a critical point — is the one you have to think about for yourself.

7. How do you actually sketch a curve — and where did Rolle and MVT come from, anyway?

The sheet's seven-step curve-sketching procedure (§9 of page 1) is the assembly line for Problems 11 and 12. It reads:

1. Identify the domain, symmetries, periods.
2. Find y' and y''.
3. Find the critical points (if they exist) and determine the extremality.
4. Determine the monotonic intervals.
5. Find the inflection points (if they exist) and determine the concavities.
6. Identify asymptotes (if they exist).
7. Plot key points (intercepts, critical points, inflection points) and asymptotes (if they exist).classwork

Stewart's §4.5 has the same procedure as an eight-item A–H checklist, with the only difference that Stewart treats intercepts as a separate step (item B) before merging it into the plot step.stewart

Each of these steps maps back to a previous chapter: Step 3 is the First Derivative Test (Chapter 5); Step 4 is Corollary 3 of MVT (Chapter 3); Step 5 is the Concavity Test and inflection-point analysis (Chapter 6). Steps 1, 2, 6, 7 are algebra and arithmetic — find the domain, take derivatives, factor, plot. The seven-step procedure is just a checklist for "which previous chapter's tools do I apply, in which order?"

Worked Example: Problem 12(a), f(x) = x³ − 3x + 1. Apply the seven steps.

• Step 1. Domain is ℝ. The function is neither even nor odd; no periodicity.
• Step 2. f'(x) = 3x² − 3 = 3(x − 1)(x + 1). f''(x) = 6x.
• Step 3. f'(x) = 0 at x = ±1. These are the critical points. By the First Derivative Test: on (−∞, −1), f'(x) > 0 (function increasing); on (−1, 1), f'(x) < 0 (function decreasing); on (1, ∞), f'(x) > 0 (function increasing). So x = −1 is a local maximum with value f(−1) = (−1)³ − 3(−1) + 1 = 3, and x = 1 is a local minimum with value f(1) = 1³ − 3 · 1 + 1 = −1.
• Step 4. Same as Step 3: increasing on (−∞, −1) and (1, ∞), decreasing on (−1, 1).
• Step 5. f''(x) = 0 at x = 0. Sign of f'': negative for x < 0 (concave down), positive for x > 0 (concave up). Concavity changes at 0, so x = 0 is an inflection point with value f(0) = 1.
• Step 6. No asymptotes (it's a polynomial).
• Step 7. Plot the local max (−1, 3), local min (1, −1), and inflection point (0, 1). The curve rises from −∞, peaks at (−1, 3), falls to (1, −1), then rises again to +∞, with the concavity switching from down to up at (0, 1).stewart

Worked Example: Problem 12(b), f(x) = x⁴ − 4x³. This one has a subtle repeated-root feature.

• Step 1. Domain ℝ; polynomial.
• Step 2. Factor: x⁴ − 4x³ = x³(x − 4). So f(x) = 0 at x = 0 (triple root) and x = 4 (simple root). f'(x) = 4x³ − 12x² = 4x²(x − 3). f''(x) = 12x² − 24x = 12x(x − 2).
• Step 3. f'(x) = 0 at x = 0 and x = 3. Sign of f': on (−∞, 0), 4x² > 0 and (x − 3) < 0, so f' < 0 (decreasing); on (0, 3), 4x² > 0 and (x − 3) < 0, so f' < 0 (still decreasing); on (3, ∞), f' > 0 (increasing). Same sign on both sides of 0 ⇒ x = 0 is not a local extremum. Sign change − to + at x = 3 ⇒ x = 3 is a local minimum with value f(3) = 81 − 108 = −27. Since f(x) ≥ 0 for x ≥ 4 and f(4) = 0, the absolute minimum on [0, ∞) is f(0) = 0 (the curve touches the x-axis at 0 without crossing — that's what the triple root means).
• Step 4. Decreasing on (−∞, 3); increasing on (3, ∞). Note: the triple root at 0 does not interrupt the monotonicity.
• Step 5. f''(x) = 0 at x = 0 and x = 2. Sign of f'': f'' > 0 on (−∞, 0) ∪ (2, ∞) (concave up); f'' < 0 on (0, 2) (concave down). Concavity changes at 0 and at 2, so both are inflection points with values f(0) = 0 and f(2) = 16 − 32 = −16.
• Step 6. No asymptotes (polynomial).
• Step 7. Plot the local minimum (3, −27), the inflection points (0, 0) and (2, −16), and the x-intercepts (0, 0) and (4, 0). The curve descends from +∞, touches the x-axis at (0, 0) (the triple root — the curve is tangent to the x-axis there but does not cross), continues descending through (2, −16) and (3, −27), then rises sharply to cross the x-axis at (4, 0) and on to +∞.

The historical arc: Rolle's Theorem was proved in 1691, but only for polynomials, and Rolle himself considered the methods of calculus fallacious at the time — his proof was algebraic, not analytic. The first rigorous modern proof of Rolle's Theorem came from Cauchy in 1823, who proved MVT in its full ε-δ form and showed Rolle's Theorem as a corollary (the opposite direction from how it is usually taught today, where Rolle's Theorem comes first and MVT is derived from it via the auxiliary function).wikipedia

The naming has its own arc. Moritz Wilhelm Drobisch (Germany, 1802–1896) first used the name "Rolle's theorem" in 1834; Giusto Bellavitis (Italy, 1803–1880) independently named it in 1846 — both were reacting to Cauchy's recasting of the original 1691 result.wikipedia The Wikipedia article notes a much earlier precursor: Parameshvara of the Kerala School (1380–1460) described a special case of MVT for inverse interpolation of the sine in his commentaries on Govindasvāmi and Bhāskara II, and Bhāskara II himself stated an early form of Rolle's Theorem in the 12th century — but without a modern formal proof.wikipedia

A final note on attribution. Some calculus textbooks (including Stewart) credit MVT to Lagrange; Stewart's slide deck says "another French mathematician, Joseph-Louis Lagrange" stated it first.stewart This attribution is widely repeated in textbooks but not directly verified from the Wikipedia source for this tutorial, which moves from Rolle 1691 directly to Cauchy 1823 without naming Lagrange. Treat the Lagrange attribution as a textbook tradition worth flagging, not as a verified historical fact.

To close the loop, here is what each chapter unlocked on the worksheet:

The worksheet's own 3×3 grid (Figure 5) is the at-a-glance reference every worked example from Chapter 5 onwards cross-checks against. The seven-step procedure (Figure 7 above) is the assembly line. And the two theorems the sheet states without proof — Rolle and MVT — turn out to be the only tools the rest of the worksheet needs.

References

1. classwork9.pdf — Calculus classwork sheet (Rolle, MVT, derivative tests, concavity, curve sketching, 12-problem set) — class handout; user-uploaded primary source. Direct link
2. OpenStax Calculus Volume 1, §§4.3–4.5 (Maxima and Minima, Mean Value Theorem, Derivatives and the Shape of a Graph) — https://openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem
3. Stewart, Calculus: Early Transcendentals (8th ed., 2016), Chapter 4 §§4.2 / 4.5, via LibreTexts adaptation — https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/04%3A_Applications_of_Differentiation/4.02%3A_The_Mean_Value_Theorem and https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/04%3A_Applications_of_Differentiation/4.05%3A_Summary_of_Curve_Sketching
4. Wikipedia: Mean value theorem; Rolle's theorem — https://en.wikipedia.org/wiki/Mean_value_theorem and https://en.wikipedia.org/wiki/Rolle%27s_theorem
5. CalcLearn: When Rolle's Theorem Fails: Three Counterexamples (Jude Wallis, 2026-08-11) — https://www.calclearn.org/theorems/rolles-theorem-counterexamples
6. Wolfram MathWorld: Rolle's Theorem (Eric W. Weisstein) — https://mathworld.wolfram.com/RollesTheorem.html