# Chapter 11: The Derivative and Graphs > Turn derivatives into a full profile of a function: locate global and local extrema, prove Rolle's and the Mean Value Theorem, and read concavity straight off the second derivative. **Canonical:** https://calculus.kitzo.work/chapter/derivative-graphs **Level:** Core **Estimated time:** 45 minutes **Exercises:** 8 (auto-marked; answers stay server-side) ## Topics - Global vs local extrema - Extreme Value Theorem - Finding global max/min - Rolle's theorem - Mean Value Theorem - Concavity & inflection points - First derivative test - Second derivative test **Why this matters** Derivatives turn a curve from a picture into a machine you can interrogate. Where a graph peaks, where it bends, where it hides a surprise — every one of those features is read off f′ and f″. This chapter is the toolkit for questioning a function hard: finding every peak and valley, proving theorems that guarantee hidden points, and judging the shape of a curve before you ever plot it. ## 1 · Global vs local extrema A **local maximum** at x = c means f(c) ≥ f(x) for all x near c — a peak of the neighbourhood. A **global (absolute) maximum** means f(c) ≥ f(x) for every x in the domain — the tallest point overall. Every global extremum in the interior of the domain is also local, but a local extremum need not be global, and a global extremum may sit on the boundary, where f′ has no reason to vanish. The **Extreme Value Theorem** guarantees that a continuous function on a closed, bounded interval [a, b] attains both a global maximum and a global minimum. Both hypotheses matter: x² on the open interval (0, 1) never attains its supremum; 1/x on (0, 1] grows without bound. ## 2 · Finding the global max and min The candidate method is mechanical and reliable on any closed interval: - Find where f′(x) = 0 or f′(x) fails to exist — the **critical points** — and keep only those inside [a, b]. - Keep the endpoints a and b as candidates as well. - Evaluate f at every candidate. The largest value is the global maximum; the smallest is the global minimum. On an open or infinite interval there may be no global extremum at all; running the candidate check and inspecting behaviour at infinity is how you discover that. **Worked example 1** ### Global extrema of f(x) = x³ − 3x on [0, 2] - **Differentiate:** f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Critical points: x = 1 and x = −1. Only x = 1 lies inside [0, 2]. - **List candidates:** x = 0, x = 1, x = 2. - **Evaluate:** f(0) = 0, f(1) = 1 − 3 = −2, f(2) = 8 − 6 = 2. - **Conclude:** global minimum −2 at x = 1; global maximum 2 at x = 2. Note the global max occurs at an endpoint, not at a critical point — skipping endpoints is the classic error here. ## 3 · Rolle's theorem and the Mean Value Theorem **Rolle's theorem:** if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f′(c) = 0. Geometrically: a curve that starts and ends level must flatten out somewhere in between. **The Mean Value Theorem (MVT)** drops the equal-height requirement. Under the same continuity and differentiability hypotheses, there exists c in (a, b) with f′(c) = (f(b) − f(a)) / (b − a) Rolle is the special case f(a) = f(b). A useful consequence: the average rate of change over [a, b] is always achieved exactly somewhere inside. This is also the engine behind error estimates for linear approximation — remember it for the next chapter. **Watch out** MVT demands differentiability on the *open* interval and continuity on the closed one. f(x) = |x| on [−1, 1] satisfies f(1) = f(−1) = 0, but no f′ vanishes — MVT is not violated, because f fails to be differentiable at 0. Likewise a single jump discontinuity anywhere in [a, b] voids the whole conclusion, even if f is smooth elsewhere. ## 4 · The second derivative and concavity f″(x) > 0 on an interval means f′ is increasing and the graph is **concave up** — it cups, like a bowl collecting water. f″(x) < 0 means **concave down** — it caps, like an umbrella. A point where f″ changes sign is a **point of inflection**: the curve switches between cupping and capping, and the tangent line crosses from one side of the curve to the other. An inflection point requires an actual sign change — a zero of f″ alone is not enough, as x⁴ at x = 0 shows. ## 5 · Classifying critical points: the two tests The **first derivative test** classifies a critical point by the sign change of f′: positive → negative means a local maximum; negative → positive means a local minimum; no sign change means a stationary inflection, like x³ at 0. The **second derivative test** uses f″ directly at c: f′(c) = 0 with f″(c) > 0 gives a local minimum; f′(c) = 0 with f″(c) < 0 gives a local maximum; f″(c) = 0 is inconclusive, and you fall back on the first derivative test. **Worked example 2** ### Classify the critical points of f(x) = 2x³ − 9x² + 12x + 1 - **Find critical points:** f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2), so x = 1 and x = 2. - **Apply f″:** f″(x) = 12x − 18. Then f″(1) = −6 < 0, so x = 1 is a local maximum with f(1) = 6; and f″(2) = 6 > 0, so x = 2 is a local minimum with f(2) = 5. - **Inflection check:** f″(x) = 0 gives x = 3/2. Since f″ is a non-constant linear function it changes sign there, so (3/2, f(3/2)) = (3/2, 11/2) is a genuine inflection point — every cubic has exactly one. **Chapter formula sheet** Extreme Value Theorem: continuous on [a, b] ⇒ global max & min exist Candidates: f′(x) = 0, f′ undefined, endpoints a, b Rolle: f(a) = f(b) ⇒ ∃ c ∈ (a, b) with f′(c) = 0 MVT: f′(c) = (f(b) − f(a)) / (b − a) f″ > 0 concave up; f″ < 0 concave down Inflection point: f″ changes sign f′(c) = 0, f″(c) < 0 ⇒ local max; f″(c) > 0 ⇒ local min ## Common mistakes - Forgetting endpoints when hunting global extrema on [a, b] — the largest value often lives there. - Assuming every f′(x) = 0 is a max or min; it can be a stationary inflection, as x³ at 0. - Calling a zero of f″ an inflection point without verifying the sign change. - Applying Rolle's theorem or MVT without checking continuity on [a, b] and differentiability on (a, b). - Reading a local maximum as the global maximum on a closed interval without comparing endpoint values. - Treating f″(c) = 0 as a dead end instead of switching to the first derivative test. ## Practice exercises 1. If f′(c) = 0 and f″(c) = −4, what can you conclude about x = c? - A. f has a local maximum at c - B. f has a local minimum at c - C. c is a point of inflection - D. f has a global minimum at c, guaranteed 2. Find the global maximum value of f(x) = x³ − 3x on the closed interval [0, 2]. 3. Find the x-coordinate of the inflection point of f(x) = x³ − 9x² + 24x. Give your answer as e.g. x=3. 4. Rolle's theorem says that if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then: - A. there is at least one c in (a, b) with f′(c) = 0 - B. f has a maximum at one of the endpoints - C. f′ is positive throughout (a, b) - D. there is exactly one c in (a, b) with f′(c) = 0 5. For f(x) = x² on [1, 3], the Mean Value Theorem guarantees a point c with f′(c) = (f(3) − f(1))/(3 − 1). Find c. 6. Using the first derivative test, if f′ changes from positive to negative at a critical point c, then c is a: - A. local minimum - B. local maximum - C. point of inflection - D. endpoint where the derivative is undefined 7. Find the local maximum value of f(x) = 2x³ − 9x² + 12x + 1. 8. For f(x) = x³ − 3x², give the x-coordinate of the inflection point as e.g. x=1. --- Practice interactively: https://calculus.kitzo.work/chapter/derivative-graphs Citation: CalculusKitzo (https://calculus.kitzo.work/). Independent study notes inspired by the standard calculus syllabus; not affiliated with Princeton University Press. See https://calculus.kitzo.work/disclaimer