On a restricted interval, the greatest and least values can occur at stationary points or at the endpoints. List every candidate, evaluate the function at each, and compare.
This lesson is part of recovering the method in a mixed calculus problem. It builds on finding stationary points.
The candidate-list method
- Differentiate and solve dy/dx = 0.
- Keep only the solutions inside the given interval.
- Evaluate the function at those points and at both endpoints.
- The largest value is the greatest, and the smallest is the least.
Worked example 1: the stationary point is outside
A profit function is P(x) = −x² + 12x − 20 for 0 ≤ x ≤ 4, with x in hundreds of units and P in RM hundreds.
P′(x) = −2x + 12 = 0 gives x = 6. This lies outside the interval [0, 4], so it is not a candidate.
The candidates are the endpoints: P(0) = −20 and P(4) = −16 + 48 − 20 = 12. The greatest profit is 12 at x = 4, which is an endpoint. A student who stopped at x = 6 would quote P(6) = 16, a value that cannot occur within the limits.
Worked example 2: a tie
Find the greatest and least values of y = x³ − 9x² + 24x for 0 ≤ x ≤ 5.
dy/dx = 3x² − 18x + 24 = 3(x − 2)(x − 4), so x = 2 or x = 4. Both are inside the interval.
| x | y = x³ − 9x² + 24x |
|---|---|
| 0 (endpoint) | 0 |
| 2 (stationary) | 8 − 36 + 48 = 20 |
| 4 (stationary) | 64 − 144 + 96 = 16 |
| 5 (endpoint) | 125 − 225 + 120 = 20 |
The greatest value is 20, reached at both x = 2 and x = 5. The least value is 0 at x = 0.
The local minimum at x = 4 has a value of 16, but it is not the least value on the interval, because the endpoint x = 0 gives a smaller one.
The mistake that costs marks
The common slip is to give a stationary point as the answer without testing the endpoints.
| Step | Wrong | Right |
|---|---|---|
| Candidates | Stationary points only | Stationary points inside the interval and both endpoints |
| Conclusion | “Maximum at x = 2” | “Greatest value is 20, at x = 2 and x = 5” |
Write the table of values, so the marker can see you checked every candidate.
Check yourself
Find the greatest and least values of f(x) = x³ − 6x² + 9x + 1 for 0 ≤ x ≤ 5.
Answer
f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), so x = 1 or x = 3.
f(0) = 1. f(1) = 1 − 6 + 9 + 1 = 5. f(3) = 27 − 54 + 27 + 1 = 1. f(5) = 125 − 150 + 45 + 1 = 21.
The greatest value is 21 at x = 5. The least value is 1, reached at both x = 0 and x = 3.
What to study next
Go on to what a negative derivative means in a model, or attempt the integrated practice set.
If you want a teacher to review your candidate tables, see online one-to-one Additional Mathematics tuition.