9 min readPathway sheet3-mark question; often part (iii) of a 4-mark case-based question

Absolute maximum and minimum on a closed interval

On a closed interval [a, b], find f at the critical points inside and at the end points. The largest is the absolute maximum, the smallest the absolute minimum.

ExistenceA function continuous on a closed interval [a, b] has an absolute maximum and an absolute minimum there

Where they occurAt critical points inside the interval or at the end points a and b

Strictly increasing f on [a, b]Absolute minimum f(a), absolute maximum f(b)

Other namesAbsolute maximum = greatest value; absolute minimum = least value

Why end points matter

A continuous function on a closed interval [a, b] always reaches a greatest value and a least value. Each one occurs either at a critical point inside the interval or at an end point. A local maximum inside may be beaten by the value at an end, so you must compare all of them. On an open interval, or on all of R, a greatest or least value may not exist.

Working rule

1
Check that f is continuous on [a, b].
2
Find f′(x) and the critical points. Keep only those inside [a, b].
3★
Find f at every critical point kept and at both end points a and b.
4
The largest of these values is the absolute maximum; the smallest is the absolute minimum.
5
Write each value with the point where it occurs.

★ key step

Example 1: a cubic on [0, 3]

Problem

Find the absolute maximum and absolute minimum values of f(x) = 2x³ − 9x² + 12x + 5 on [0, 3].

Working

f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2)
Critical points: x = 1 and x = 2, both inside [0, 3].
f(0) = 5
f(1) = 2 − 9 + 12 + 5 = 10
f(2) = 16 − 36 + 24 + 5 = 9
f(3) = 54 − 81 + 36 + 5 = 14
Largest value 14; smallest value 5.

Answer

Absolute maximum 14 at x = 3; absolute minimum 5 at x = 0. The local maximum value 10 at x = 1 is not the absolute maximum.

Example 2: a trigonometric function on [0, π]

Problem

Find the absolute maximum and minimum values of f(x) = x + 2 cos x on [0, π].

Working

f′(x) = 1 − 2 sin x
f′(x) = 0 gives sin x = 1/2, so x = π/6 or x = 5π/6. Both lie in [0, π].
f(0) = 0 + 2 × 1 = 2
f(π/6) = π/6 + 2 × (√3/2) = π/6 + √3 ≈ 0.524 + 1.732 = 2.256
f(5π/6) = 5π/6 − √3 ≈ 2.618 − 1.732 = 0.886
f(π) = π − 2 ≈ 1.142
Compare (π ≈ 3.1416, √3 ≈ 1.732): the largest is π/6 + √3 and the smallest is 5π/6 − √3.

Answer

Absolute maximum π/6 + √3 at x = π/6; absolute minimum 5π/6 − √3 at x = 5π/6.

Local extreme vs absolute extreme on [a, b]

Local maximum or minimumFound at a critical point inside the interval using the first or second derivative test. There can be several.

Absolute maximum or minimumFound by comparing f at the critical points and the end points. There is exactly one absolute maximum value and one absolute minimum value on [a, b].

An end point can give the absolute maximum without being a point of local maxima.

Traps

Do not forget the end points. This is the most common lost mark.
Drop critical points that lie outside [a, b], and say why.
You do not need the second derivative test here; just compare values.
If f′(x) exists and is never zero inside [a, b], f is monotonic and the extreme values are f(a) and f(b).
A point where f′ does not exist is still a critical point: f(x) = ∛(x²) has f′(x) = 2/(3∛x), undefined at x = 0.

In real life

A school canteen can cook between 50 and 200 plates of idli a day. The most profitable number may be at a turning point or at one of these limits, so both must be checked.

Exam angle

1 marks: “A function is strictly decreasing on [1, 5]. Where does it take its absolute maximum value?” At x = 1, the left end point, because the values keep falling as x increases.

2 marks: “Find the absolute maximum and minimum values of f(x) = x² − 2x on [0, 3].” f′(x) = 2x − 2 = 0 at x = 1. f(0) = 0, f(1) = −1, f(3) = 3. Absolute maximum 3 at x = 3; absolute minimum −1 at x = 1.

3 marks: “Find the absolute maximum and minimum values of f(x) = x³ − 3x² − 9x + 4 on [−2, 4].” f′(x) = 3(x − 3)(x + 1), critical points −1 and 3. f(−2) = 2, f(−1) = 9, f(3) = −23, f(4) = −16. Absolute maximum 9 at x = −1; absolute minimum −23 at x = 3.

Remember it

Ends and bends. Check both end points and every bend (critical point) inside.