AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/5/5-5)
Unit 5 · Topic 5.5
5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
To find the absolute maximum and minimum of a continuous function on a closed interval, test every candidate: the critical points inside the interval and both endpoints. Evaluate f at each, and the largest and smallest outputs win. This is the Candidates Test, also called the Closed Interval Method.
Key terms
- Candidates Test
- Closed Interval Method
- absolute maximum
- absolute minimum
- endpoints
Why the candidates are enough
By the Extreme Value Theorem, a continuous function on [a, b] has an absolute max and min. Each one happens either at an endpoint or at an interior point. An interior extreme is also a relative extreme, and relative extremes only happen at critical points. So the max and min must be at a critical point or an endpoint. Nowhere else needs checking.
The steps
The routine is short:
- Confirm f is continuous on [a, b].
- Find all critical points of f in the open interval (a, b). Throw out any outside it.
- Evaluate f at each critical point and at x = a and x = b.
- The largest value is the absolute max; the smallest is the absolute min.
Show it as a table
A table of candidates is clear and is accepted as justification on free response. Label it, include every candidate, and then state your conclusion in a sentence. A tie is allowed: the max or min can happen at more than one x-value.
| x | f(x) | Note |
|---|---|---|
| a (left endpoint) | f(a) | Candidate |
| critical point c | f(c) | Candidate |
| b (right endpoint) | f(b) | Candidate |
When the function comes from an integral
In later units, f is sometimes defined by an integral, and you're given f′ and one value of f. The Candidates Test still works, but finding each f(x) takes integrals (Unit 6 and Unit 8). The logic is the same.
Open intervals and missing endpoints
The Candidates Test only works on a closed interval where f is continuous, because EVT is what guarantees the max and min exist. On an open or infinite interval, there may be no max or min at all: f(x) = x² on (1, 3) gets close to 1 and 9 but never reaches either.
For those intervals, look at the critical points with a derivative test, and use the one-critical-point idea from 5.7 when it applies. You can also study the behavior near the ends with limits.
Using a calculator
On calculator-active questions, the calculator can find critical points by solving f′(x) = 0, or by finding where the graph of f′ crosses the x-axis. You still have to list the candidates and compare their values in your written answer. Store each critical point in your calculator so you don't lose accuracy when you plug it into f.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
A polynomial on a closed interval
Find the absolute maximum and minimum values of f(x) = x³ − 3x² + 1 on [−1, 4].
Show the solutionHide the solution
- Step 1: f is a polynomial, so it's continuous on [−1, 4].
- Step 2: f′(x) = 3x² − 6x = 3x(x − 2). Critical points: x = 0 and x = 2, both inside (−1, 4).
- Step 3: Candidates: f(−1) = −1 − 3 + 1 = −3; f(0) = 1; f(2) = 8 − 12 + 1 = −3; f(4) = 64 − 48 + 1 = 17.
- Step 4: Largest: 17. Smallest: −3, which occurs twice.
Answer: Absolute maximum 17 at x = 4. Absolute minimum −3 at both x = −1 and x = 2.
- Example 2
A trig function
Find the absolute extrema of f(x) = x − 2 sin x on [0, π].
Show the solutionHide the solution
- Step 1: f is continuous on [0, π].
- Step 2: f′(x) = 1 − 2 cos x = 0 when cos x = 1/2. In (0, π), that's x = π/3.
- Step 3: Candidates: f(0) = 0; f(π/3) = π/3 − 2(√3/2) = π/3 − √3 ≈ −0.685; f(π) = π − 0 = π.
- Step 4: Compare: π is largest, π/3 − √3 is smallest.
Answer: Absolute maximum π at x = π. Absolute minimum π/3 − √3 ≈ −0.685 at x = π/3.
- Example 3
Trap: a critical point outside the interval
Find the absolute extrema of f(x) = x³ − 12x on [0, 3].
Show the solutionHide the solution
- Step 1: f′(x) = 3x² − 12 = 3(x − 2)(x + 2). Critical points: x = 2 and x = −2.
- Step 2: x = −2 is outside [0, 3], so it isn't a candidate. Including it would add f(−2) = 16, which isn't a value f takes on this interval.
- Step 3: Candidates: f(0) = 0; f(2) = 8 − 24 = −16; f(3) = 27 − 36 = −9.
Answer: Absolute maximum 0 at x = 0. Absolute minimum −16 at x = 2.
Common mistakes
- Forgetting the endpoints. The absolute max is often at an endpoint.
- Including critical points that lie outside the interval.
- Answering with the x-value when the question asks for the maximum value (a y-value), or the reverse.
On the exam
- “Find the absolute maximum value of f on [a, b]. Justify your answer.” A candidates table plus a concluding sentence is the standard full-credit response.
- If the interval is open or infinite, the Candidates Test doesn't apply directly. Use the one-critical-point idea from 5.7 or analyze the behavior near the ends.
Connected topics
Videos
Check yourself
4 questions on 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema. Pick an answer to see if you got it, and why.
What is the absolute minimum value of f(x) = x³ − 6x² + 9x + 1 on the closed interval [−1, 4]?
Let f(x) = e^(sin x) − x/2. What is the absolute maximum value of f on the closed interval [0, 6]?
What is the absolute maximum value of f(x) = x − 2ln x on the closed interval [1, e²] ?
Let f(x) = 3x^(2/3) − 2x. What are the absolute maximum and absolute minimum values of f on the closed interval [−1, 8] ?
0 of 4 answered