On a restricted interval, the greatest and least values of a function occur either at a stationary point inside the interval or at an endpoint. Calculate the function at each candidate and compare.
This lesson extends classifying stationary behaviour and prepares you for the domain reasoning in interpreting an optimum. It sits within stationary points and optimisation.
Why can the edge win?
A turning point only tells you that the curve changes direction there. It says nothing about what happens further along. If the curve is still climbing when the interval stops, the highest value is at the end.
Think of walking over a small hill on a path that then continues uphill. The summit of the small hill is a local maximum, but the highest place on the whole path is the far end.
How to find the greatest and least values, step by step
- Find the stationary x-values with dy/dx = 0.
- Keep only those inside the interval. Discard the others.
- Calculate y at each kept stationary x-value and at both endpoints.
- List the values in a short table.
- Pick the largest and smallest and state where each occurs.
Worked example
Find the greatest and least values of y = x³ − 6x² + 9x + 2 for 0 ≤ x ≤ 5.
Step 1, stationary points: dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3), so x = 1 or x = 3. Both lie inside the interval.
Step 2, evaluate all candidates:
| x | y |
|---|---|
| 0 (endpoint) | 2 |
| 1 (stationary) | 1 − 6 + 9 + 2 = 6 |
| 3 (stationary) | 27 − 54 + 27 + 2 = 2 |
| 5 (endpoint) | 125 − 150 + 45 + 2 = 22 |
Step 3, compare: the largest value in the table is 22 and the smallest is 2.
Answer: the greatest value is 22, at x = 5, and the least value is 2, at x = 0 and x = 3.
The local maximum of 6 at x = 1 is not the greatest value on this interval.
The mistake to watch for
A common slip is to compare only the stationary values.
Mistaken answer: greatest value 6, least value 2, because those are the stationary values.
The student never tested x = 5, where the curve is much higher.
The correction is to treat the endpoints as candidates every time the question gives an interval such as 0 ≤ x ≤ 5. A table of x and y for all candidates takes a minute and prevents this error.
Check yourself
Try these without a calculator, then open each answer.
1. Find the greatest and least values of y = x² − 4x + 1 for 0 ≤ x ≤ 5.
Show answer
dy/dx = 2x − 4 = 0 gives x = 2, y = 4 − 8 + 1 = −3. Endpoints: x = 0 gives y = 1 and x = 5 gives y = 25 − 20 + 1 = 6.
Greatest 6 (at x = 5), least −3 (at x = 2)
2. Find the greatest and least values of y = x³ − 3x for −2 ≤ x ≤ 3.
Show answer
dy/dx = 3x² − 3 = 0 gives x = 1 or x = −1, both inside the interval. y(1) = −2 and y(−1) = 2. Endpoints: y(−2) = −8 + 6 = −2 and y(3) = 27 − 9 = 18.
Greatest 18 (at x = 3), least −2 (at x = 1 and x = −2)
3. Find the greatest and least values of y = 12x − x² for 0 ≤ x ≤ 4.
Show answer
dy/dx = 12 − 2x = 0 gives x = 6, which is outside the interval, so ignore it. Endpoints: y(0) = 0 and y(4) = 48 − 16 = 32.
Greatest 32 (at x = 4), least 0 (at x = 0)
Where this leads next
Next, interpret an optimum with units and a domain to write the answer in context. The non-calculator working trainer is useful for checking the endpoint arithmetic.
If you keep forgetting the edge cases under time pressure, one-to-one Additional Mathematics tuition can build a checking routine around your own working.