AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-4)
Unit 1 · Topic 1.4
1.4 Polynomial Functions and Rates of Change
Polynomials are the sums of power terms like 3x⁴ − x + 7. This topic is about the high and low points of their graphs (relative and absolute extrema) and the points where they change concavity (points of inflection), which you find from how the rate of change behaves.
Key terms
- degree
- leading coefficient
- relative (local) extremum
- absolute (global) extremum
- point of inflection
What counts as a polynomial
A polynomial function has the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₂x² + a₁x + a₀, where n is a positive integer, every coefficient is a real number, and aₙ ≠ 0.
The degree is n, the highest power. The leading term is aₙxⁿ and the leading coefficient is aₙ. For p(x) = −2x³ + 5x − 1, the degree is 3, the leading term is −2x³, and the leading coefficient is −2. A nonzero constant like p(x) = 6 is a polynomial of degree 0.
Powers must be whole numbers. Functions like √x or 1/x are not polynomials.
Relative and absolute extrema
A relative (local) maximum is an output that is greater than or equal to the outputs at all nearby inputs; a relative minimum is less than or equal to the nearby outputs. For a polynomial, these happen where the graph switches between increasing and decreasing.
If the domain is restricted to an interval and an endpoint is included, that endpoint is also a relative maximum or minimum, because the graph stops there.
An absolute (global) maximum is the greatest output on the whole domain, and an absolute minimum is the least. A relative extremum is global only if it beats every other output.
Two useful facts: between any two distinct real zeros of a nonconstant polynomial there is at least one relative maximum or minimum, and a polynomial of degree n has at most n − 1 turning points.
Even degree means a global extremum
A polynomial of even degree has both ends pointing the same way, so it always has a global maximum or a global minimum. With a positive leading coefficient, both ends go up and there is a global minimum. With a negative one, there is a global maximum.
A quadratic's global extremum is at its vertex. For p(x) = a(x − h)² + k, the vertex is (h, k).
On the whole real line, an odd-degree polynomial has no global maximum or minimum, because one end goes up forever and the other goes down forever.
Points of inflection
A point of inflection marks a switch in how the rate of change behaves: the rate stops growing and starts shrinking, or stops shrinking and starts growing. On the graph, it's where the concavity flips, from concave up to concave down or the other way.
On a cubic, there is exactly one point of inflection. It sits halfway between the two turning points when the cubic has them, because a cubic's graph is symmetric about its inflection point.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Extrema on a restricted domain
Let p(x) = x³ − 6x² + 9x + 1 on the domain 0 ≤ x ≤ 4. A graphing calculator shows that p has a relative maximum at (1, 5) and a relative minimum at (3, 1) inside the interval. Find the absolute maximum and minimum values of p on this domain, and the point of inflection.
Show the solutionHide the solution
- Step 1: On a restricted domain, the candidates for absolute extrema are the turning points and the endpoints.
- Step 2: Evaluate the endpoints: p(0) = 1 and p(4) = 64 − 96 + 36 + 1 = 5.
- Step 3: Compare all candidates: p(0) = 1, p(1) = 5, p(3) = 1, p(4) = 5.
- Step 4: The largest output is 5 and the smallest is 1. Each happens twice.
- Step 5: A cubic's inflection point is halfway between its turning points, at x = 2, and p(2) = 8 − 24 + 18 + 1 = 3.
Answer: Absolute maximum 5 (at x = 1 and x = 4); absolute minimum 1 (at x = 0 and x = 3); point of inflection (2, 3).
- Example 2
Counting guaranteed extrema
A polynomial q has real zeros only at x = −3, x = 1 and x = 4. What is the least number of relative extrema q must have between x = −3 and x = 4?
Show the solutionHide the solution
- Step 1: Between any two distinct real zeros of a polynomial there is at least one relative maximum or minimum.
- Step 2: The zeros split [−3, 4] into two gaps: from −3 to 1 and from 1 to 4.
- Step 3: Each gap needs at least one extremum.
Answer: At least 2 relative extrema: one between −3 and 1, and one between 1 and 4.
- Example 3
Trap: which global extremum?
A student says, “Every degree-4 polynomial has a global maximum.” Is this true for p(x) = 2x⁴ − 5x² + 1?
Show the solutionHide the solution
- Step 1: Even degree guarantees a global maximum or a global minimum, not specifically a maximum.
- Step 2: The leading coefficient is 2, which is positive, so both ends of the graph go up: as x → ±∞, p(x) → ∞.
- Step 3: Outputs grow without bound, so there is no greatest output. The graph does have a lowest point, a global minimum.
Answer: False for this polynomial. It has a global minimum but no global maximum.
Common mistakes
- Forgetting the endpoints when the domain is restricted. An endpoint can be a relative extremum and can be the absolute one.
- Assuming an even-degree polynomial has a global maximum. It depends on the sign of the leading coefficient.
- Calling any point where the graph is steep a point of inflection. It must be where concavity actually changes.
- Giving an extremum as an x-value when the question asks for the maximum value. The maximum value is the output, not the input where it happens.
On the exam
- With a calculator, you may need to find relative extrema; report coordinates to three decimal places.
- Questions often describe a graph in words or give a table and ask where a relative extremum or point of inflection must occur. Justify with “switches from increasing to decreasing” or “rate of change switches from increasing to decreasing.”
Connected topics
Videos
Check yourself
4 questions on 1.4 Polynomial Functions and Rates of Change. Pick an answer to see if you got it, and why.
Let p(x) = x⁴ − 5x² + x + 2. What is the absolute minimum value of p?
For a polynomial function f, the rate of change of f is increasing for x < 2 and decreasing for x > 2. Which of the following must be true?
Let f(x) = x³ − 6x² + 9x + 1. On which of the following intervals is f decreasing?
A polynomial function f is increasing on (−∞, −1), decreasing on (−1, 4) and increasing on (4, ∞). Also, f(−1) = 6 and f(4) = −20. Which of the following is true?
0 of 4 answered