Maxima and minima: local and absolute values
A local maximum is a peak compared with nearby points; the absolute maximum is the greatest value on the whole interval. Minima work the same way.
Maximum and minimum value
Let f be defined on an interval I. f has a maximum value in I if some point c in I gives f(c) ≥ f(x) for every x in I. Then f(c) is the maximum value and c is a point of maximum. The minimum value is defined in the same way with ≤. Either one is an extreme value. A function may have no maximum or no minimum: f(x) = x² on R has minimum value 0 at x = 0 but no maximum value.
Local maxima, local minima and critical points
Think of a graph as a line of hills. A hilltop is higher than the points just around it, even if another hill is taller. Such a point is a point of local maxima. The bottom of a valley is a point of local minima. At a smooth hilltop or valley the graph is flat for an instant (its slope is 0), so f′(c) = 0. At a sharp corner, such as x = 0 for f(x) = |x|, f′ does not exist. Both kinds of point are critical points. But f′(c) = 0 does not always mean a maximum or minimum: f(x) = x³ has f′(0) = 0 and keeps rising.
Local maximum vs absolute maximum
Example: f(x) = x + 1/x has local maximum value −2 at x = −1 and local minimum value 2 at x = 1.
Values of some functions on R
| Function | Maximum value | Minimum value |
|---|---|---|
| f(x) = x² + 2 | None | 2, at x = 0 |
| f(x) = 10 − (x + 3)² | 10, at x = −3 | None |
| f(x) = |x − 3| + 4 | None | 4, at x = 3 |
| f(x) = 3 cos x + 2 | 5 | −1 |
| f(x) = x³ | None | None |
Example 1: by completing the square
Find the maximum and minimum values, if any, of f(x) = 4x² − 4x + 7.
- Complete the square: 4x² − 4x + 7 = (4x² − 4x + 1) + 6
- = (2x − 1)² + 6
- (2x − 1)² ≥ 0 for every real x, so f(x) ≥ 6.
- f(x) = 6 when 2x − 1 = 0, that is x = 1/2.
- (2x − 1)² can be as large as we like, so f has no greatest value.
Minimum value 6 at x = 1/2. There is no maximum value.
Example 2: a function with a modulus
Find the maximum and minimum values of f(x) = 7 − |cos 3x|, x in R.
- For every x, 0 ≤ |cos 3x| ≤ 1.
- Multiply by −1: −1 ≤ −|cos 3x| ≤ 0.
- Add 7: 6 ≤ 7 − |cos 3x| ≤ 7.
- f(x) = 7 when cos 3x = 0, for example at x = π/6.
- f(x) = 6 when |cos 3x| = 1, for example at x = 0.
Maximum value 7, minimum value 6.
Chennai's temperature has a highest value each afternoon. Over a month there are many daily peaks (local maxima) but only one hottest moment of the month (the absolute maximum).
1 marks: “Find the minimum value of f(x) = x² + 4x + 9.” x² + 4x + 9 = (x + 2)² + 5 ≥ 5. Minimum value 5, at x = −2.
2 marks: “Find the critical points of f(x) = x³ − 27x + 10.” f′(x) = 3x² − 27 = 3(x − 3)(x + 3). f′(x) = 0 gives x = 3 and x = −3. These are the critical points.
2 marks: “Find the maximum and minimum values, if any, of f(x) = 2x + 3 on the open interval (0, 1).” f is strictly increasing, and its values fill (3, 5). f comes as close as we like to 3 and to 5 but never reaches them, so it has neither a maximum nor a minimum value.
Local is a hilltop; absolute is the tallest hill. Check corners and end points too.