Lesson 03: Polynomial Functions — Extrema, Concavity & Inflection

Warm-Up

Picture a roller coaster from the side. Every crest is a spot where you were rising and then start falling; every dip, the reverse. But there are also moments between crest and dip — mid-drop — where the track stops steepening and starts flattening. Riders feel those moments in their stomachs.

Crests and dips are extrema. The mid-drop moment is a point of inflection. Polynomials are the smoothest roller coasters in mathematics, and this lesson maps their geometry.

Core Concept

Polynomial functions

A polynomial function has the form

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀,   aₙ ≠ 0

n (a nonnegative integer) is the degree; aₙ is the leading coefficient. Linear and quadratic functions are degrees 1 and 2; now we go higher.

Local vs. global extrema

• A local (relative) maximum occurs at x = c if f(c) ≥ f(x) for all x near c — the graph switches from increasing to decreasing there.
• A local minimum: switches from decreasing to increasing.
• A global (absolute) maximum/minimum is the greatest/least output over the entire domain.

Key structural facts for polynomials:

1. Even degree → guaranteed global extremum. Both ends of an even-degree polynomial point the same way (both up if aₙ > 0, both down if aₙ < 0), so it has a global minimum (if up) or global maximum (if down) — but not both.
2. Odd degree → no global extrema. The ends run to +∞ and −∞, so every extremum of an odd-degree polynomial is local only.
3. Between any two distinct real zeros, there is at least one local extremum. The graph touches zero twice, so somewhere in between it must turn around. (This is the CED's version of Rolle's theorem — no calculus needed, just the picture.)
4. A polynomial of degree n has at most n − 1 local extrema (at most n − 1 turning points).

Points of inflection

A point of inflection is a point where the graph changes concavity — where the rate of change stops increasing and starts decreasing, or vice versa.

• Concave up → concave down at x = c: the rate of change was increasing, now decreasing; the graph crosses from ∪-bend to ∩-bend.
• At an inflection point, the function itself may be rising or falling — inflection is about the bend, not the direction.

[GRAPH: Cubic-like curve for f with a local maximum at (−1, 4), a point of inflection at (1, 1) marked with a square marker, and a local minimum at (3, −2). Left of x = −1: increasing, concave down. Between −1 and 1: decreasing, concave down (falling, steepening). Between 1 and 3: decreasing, concave up (falling, flattening). Right of 3: increasing, concave up. Annotation at (1,1): "inflection: rate of change changes from decreasing to increasing".]

A degree-n polynomial has at most n − 2 points of inflection; an odd-degree polynomial (degree ≥ 3) always has at least one.

The full vocabulary picture

For the graph above, an AP-grade description of (−1, 3):

"f is decreasing on the interval (−1, 3) because outputs fall as inputs increase. f is concave down on (−1, 1) — its rate of change is decreasing — and concave up on (1, 3) — its rate of change is increasing. The point (1, 1) is a point of inflection because the concavity changes there."

Every sentence names a behavior and gives the rate-of-change reason. That's the FRQ 4 formula: claim + because-clause about rates of change.

Calculator note 📱

On calc-active items, use your grapher's minimum/maximum commands rather than tracing. Set a sensible window first (zeros and extrema of AP polynomials usually live within −10 ≤ x ≤ 10). Report at least three decimal places unless told otherwise — that's the AP's standard accuracy expectation.

Worked Examples

Example 1 (easy) — Classify the extrema 🚫 No-Calc

Problem: p is a degree-4 polynomial with negative leading coefficient and three local extrema. Which are guaranteed global?

Solution: Even degree + negative leading coefficient → both ends fall to −∞ → p has a global maximum and no global minimum. With three local extrema (max–min–max pattern for down-pointing quartics), the global maximum is the larger of the two local maxima. The local minimum is local only.

Interpretation: End behavior (Lesson 5 in full) decides which extremum can be global before you compute anything.

Example 2 (medium) — Extrema between zeros 🚫 No-Calc

Problem: f(x) = x(x − 2)(x + 3). Without graphing, how many local extrema does f have, and where (roughly)?

Strategy: Count zeros; apply the between-zeros guarantee; cap with degree.

Solution: Zeros at −3, 0, 2 — three distinct real zeros. Between −3 and 0: at least one local extremum. Between 0 and 2: at least one more. Degree 3 allows at most 3 − 1 = 2. So exactly two: one in (−3, 0), one in (0, 2). Leading coefficient positive and degree odd → rises to the right → the first is a local max, the second a local min.

Interpretation: "At least one between zeros" + "at most n − 1 total" often pins the count exactly.

Example 3 (medium) — Reading inflection from rate-of-change data 🚫 No-Calc

Problem: Equally spaced values of g:

Where do the data suggest a point of inflection?

Solution: First differences: 7, 5, 3, 3, 5. The rates decrease (7 → 5 → 3) then increase (3 → 3 → 5). The switch happens around x = 3–4, so the data suggest g changes from concave down to concave up — a point of inflection near x = 3.5 (any answer locating it in [3, 4] with the rate-reversal reason is fine).

Interpretation: Inflection from a table = find where first differences reverse trend.

Example 4 (AP-style) — Calculator extremum in context 📱 Calc-Active

Problem: The value of one share of an investment fund is modeled by V(t) = 0.1t³ − 1.8t² + 8t + 20 dollars, for 0 ≤ t ≤ 14 days. Find the maximum share value during this period.

Strategy: Graph on [0, 14]; use the maximum finder — then check the endpoints.

Solution: The calculator's maximum command finds a local maximum of V ≈ 30.503 dollars at t ≈ 2.945 days. But the graph turns back up after a local minimum near t ≈ 9.055, so check the right endpoint:

V(14) = 0.1(2744) − 1.8(196) + 8(14) + 20
      = 274.4 − 352.8 + 112 + 20 = 53.6

The endpoint beats the interior local max: on [0, 14], the maximum share value is $53.60 at t = 14, since 53.6 > 30.503.

Interpretation: On a restricted domain, always check endpoints against interior extrema. The calculator's "maximum" finds local maxima only — the exam sets this trap deliberately.

Common Mistakes

1. "Highest point on my screen" = global max. Your window can hide behavior; restricted domains have endpoint candidates. Fix: compare interior extrema against endpoint values, and know end behavior before trusting a window.
2. Calling every extremum "the maximum." Specify local vs. global; the AP language distinction is graded. Fix: local = beats its neighbors; global = beats everything.
3. Placing inflection points at extrema. Extrema are where direction changes; inflections are where bend changes. They're usually different points. Fix: separate checklists.
4. Assuming degree n means exactly n − 1 turning points. It's at most. x³ has degree 3 and zero turning points (but still an inflection at the origin). Fix: "at most," and use zeros to force minimums.
5. Reporting calculator answers to 1 decimal. AP expects three decimal places (or exact). Write t ≈ 9.107, not 9.1. Points are lost to rounding every year.

Practice Problems

FRQ Practice — Task Model: Communicating about Functions (FRQ 4 style) 🚫 No-Calc

The graph of the polynomial g passes through the points (−2, 0), (0, 3), (2, 0), and (5, −6), with a local maximum at (0, 3), a point of inflection at (2, 0), and a local minimum at (4, −7). The degree of g is 4 and its leading coefficient is positive.

(a) (i) On what open intervals is g decreasing? (ii) Give a reason for your answer based on the definition of decreasing.

(b) (i) On what open interval(s) shown is the rate of change of g decreasing? (ii) Explain how the point of inflection at (2, 0) relates to the behavior of the rate of change of g.

(c) (i) Does g have a global minimum? (ii) Explain your reasoning using the degree, leading coefficient, and the given points.

Model Response & Rubric (6 points)

(a) [2 pts] (i) [1 pt] g is decreasing on (0, 4). (End behavior: with degree 4 and positive leading coefficient, g rises on both ends; given the listed extrema, g decreases only between the local max at x = 0 and the local min at x = 4.) (ii) [1 pt] For any a < b in (0, 4), g(a) > g(b): outputs fall as inputs increase, which is the definition of decreasing.

(b) [2 pts] (i) [1 pt] On the interval shown, the rate of change of g is decreasing on (−2, 2): the graph is concave down there (it bends like ∩ around the local maximum at x = 0), and concave down means the rate of change is decreasing. (ii) [1 pt] At (2, 0) the graph changes from concave down to concave up: the rate of change of g changes from decreasing to increasing there, which is what defines a point of inflection.

(c) [2 pts] (i) [1 pt] Yes, g has a global minimum. (ii) [1 pt] g has even degree (4) and positive leading coefficient, so lim x→−∞ g(x) = ∞ and lim x→∞ g(x) = ∞: both ends rise without bound. A polynomial whose ends both rise must attain a least value, and that least value occurs at one of its local minima. (Identifying which local minimum is global would require the full graph; the guarantee itself earns the point.)