MTH1020 · Analysis of Change
Applications of Differentiation: Extrema & Concavity
Week 8 of Monash MTH1020 Analysis of Change turns the derivative into a tool for reading the shape of a graph. It covers increasing/decreasing behaviour from the sign of f′, the location and classification of local extrema by the first and second derivative tests, concavity and points of inflection from f″, and the finding of absolute extrema — including the Extreme Value Theorem on a closed interval. These techniques are examined in the mid-semester tests and the final, and they feed directly into the Week-9 curve-sketching and optimisation work.
What this chapter covers
- 01Increasing/decreasing from the first derivative: f′ > 0 ⇒ increasing, f′ < 0 ⇒ decreasing, f′ = 0 ⇒ stationary
- 02Local maxima and minima; Fermat's theorem: a differentiable local extremum has f′ = 0
- 03Critical points: where f′ = 0 or f′ does not exist — the only candidates for local extrema
- 04First derivative test: sign of f′ changing +→− gives a local max, −→+ a local min, no change neither
- 05Concavity from the second derivative: f″ > 0 ⇒ concave up, f″ < 0 ⇒ concave down
- 06Points of inflection: where concavity changes (f″ = 0 or undefined, and actually changes sign)
- 07Second derivative test: at f′(c) = 0, f″(c) > 0 ⇒ min, f″(c) < 0 ⇒ max, f″(c) = 0 inconclusive
- 08Absolute extrema and the Extreme Value Theorem: a continuous function on [a, b] attains a max and min
Classifying the turning points and inflection of a cubic
- +1Find the critical points. f′(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1). Setting f′(x) = 0 gives x = 1 and x = −1 (f′ exists everywhere, so these are the only critical points).
- +1Classify with the second derivative test. f″(x) = 6x. At x = 1, f″(1) = 6 > 0, so x = 1 is a local minimum; at x = −1, f″(−1) = −6 < 0, so x = −1 is a local maximum.
- +1Evaluate the extreme values. f(1) = 1 − 3 = −2 and f(−1) = −1 + 3 = 2. So the local minimum is (1, −2) and the local maximum is (−1, 2).
- +1Locate the inflection point. f″(x) = 6x = 0 at x = 0, and f″ changes sign from negative (x < 0, concave down) to positive (x > 0, concave up) there, so (0, f(0)) = (0, 0) is a point of inflection.
Key terms
- Critical point
- A point where f′(c) = 0 or f′(c) does not exist. Local extrema can only occur at critical points, though not every critical point is an extremum.
- First derivative test
- Classifies a critical point by the sign change of f′ across it: +→− gives a local maximum, −→+ a local minimum, no change gives neither.
- Concavity
- Concave up where f″ > 0 (the graph curves upward, lying above its tangents) and concave down where f″ < 0. Determined by the second derivative.
- Point of inflection
- A point where concavity changes. If f″ exists there it equals 0, but a sign change of f″ — not merely f″ = 0 — is required.
- Second derivative test
- At a critical point with f′(c) = 0: f″(c) > 0 ⇒ local minimum, f″(c) < 0 ⇒ local maximum, f″(c) = 0 ⇒ inconclusive (fall back on the first derivative test).
- Extreme Value Theorem
- A function continuous on a closed interval [a, b] attains both an absolute maximum and an absolute minimum on that interval. Continuity and closedness are both essential.
Applications of Differentiation: Extrema & Concavity FAQ
What's the difference between the first and second derivative tests?
Both classify a critical point, but they use different information. The first derivative test checks the sign of f′ on each side of the point: it always works, even when f′ does not exist there. The second derivative test evaluates f″ at the point: f″ > 0 means a minimum, f″ < 0 a maximum — quick when f″ is easy to compute, but inconclusive when f″(c) = 0, in which case you fall back on the first derivative test.
Does f″ = 0 always mean there's a point of inflection?
No — this is a classic trap. A point of inflection needs concavity to actually change, i.e. f″ must change sign there. The function f(x) = x⁴ has f″(0) = 0, but f″ = 12x² ≥ 0 on both sides, so concavity never changes and there is no inflection at the origin. Always confirm the sign change of f″, not just that it hits zero.
How do I find the absolute maximum on a closed interval?
By the Extreme Value Theorem, a continuous function on [a, b] does attain an absolute max and min, and they occur either at an interior critical point or at an endpoint. So the method is: find all critical points inside (a, b), evaluate f there and at both endpoints a and b, then compare the values — the largest is the absolute maximum and the smallest the absolute minimum. On open intervals or all of ℝ you instead study limits at the ends.
Can Sia help me classify turning points?
Yes. Sia can walk through finding f′ and f″, applying the first or second derivative test, and checking a sign change for an inflection — step by step and in the Monash write-up style. It explains the method and checks your reasoning on your own practice questions; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
Drill a fixed shape-analysis routine: compute f′, solve f′ = 0 for critical points, classify with whichever test is cleaner, then compute f″ = 0 and check for a genuine sign change to locate inflections. Always report extrema as coordinate points and state the concavity on each interval, because Monash marks the interpretation as well as the algebra. Keep two traps in view — f″ = 0 without a sign change is not an inflection, and on a closed interval you must also test the endpoints for absolute extrema. This chapter is examined in the mid-semester tests and the final, and it is the direct set-up for Week-9 curve sketching and optimisation, so master the routine now. Ask Sia to generate fresh functions and check your classification line by line.
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