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Maxima and Minima

A function has a local maximum at a point if its value is greater than or equal to nearby values, and a local minimum if its value is less than or equal to nearby values.

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Student-friendly explanation

Possible points of maxima or minima occur at critical points. The first derivative test checks the sign change of f'(x) around the point. The second derivative test uses f''(a): if f'(a) = 0 and f''(a) < 0, there is a local maximum; if f''(a) > 0, there is a local minimum. If f''(a) = 0, the test is inconclusive.

How to write this in exams

  1. 1

    Start with the exact idea

    A function has a local maximum at a point if its value is greater than or equal to nearby values, and a local minimum if its value is less than or equal to nearby values.

  2. 2

    Then show how to use it

    Find f'(x); solve f'(x) = 0 and list critical points in the domain; use first derivative sign change or second derivative test; calculate f(a) for each confirmed extremum; state whether it is local maximum or local minimum.

  3. 3

    Add one concrete example

    For f(x) = x^2 - 4x + 5, f'(x) = 2x - 4, so x = 2. Since f''(x) = 2 > 0, f has a local minimum at x = 2, and the minimum value is f(2) = 1.

  4. 4

    Avoid this incomplete answer

    Using f''(a) > 0 as maximum and f''(a) < 0 as minimum, which reverses the second derivative test.

Definition

A function has a local maximum at a point if its value is greater than or equal to nearby values, and a local minimum if its value is less than or equal to nearby values.

Example

For f(x) = x^2 - 4x + 5, f'(x) = 2x - 4, so x = 2. Since f''(x) = 2 > 0, f has a local minimum at x = 2, and the minimum value is f(2) = 1.

Rule to remember

First derivative test: f' changes from positive to negative at x = a gives local maximum; f' changes from negative to positive gives local minimum. Second derivative test: if f'(a) = 0 and f''(a) < 0, local maximum; if f''(a) > 0, local minimum; if f''(a) = 0, use another test.

Memory hook

Positive second derivative opens upward, so it gives a minimum; negative second derivative opens downward, so it gives a maximum.

Examples and method

Worked example

Find local extrema of f(x) = x^3 - 3x^2 - 9x + 5. f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1), so critical points are x = -1 and x = 3. f''(x) = 6x - 6. At x = -1, f''(-1) = -12 < 0, so local maximum; f(-1) = -1 - 3 + 9 + 5 = 10. At x = 3, f''(3) = 12 > 0, so local minimum; f(3) = 27 - 27 - 27 + 5 = -22. Thus local maximum value is 10 at x = -1 and local minimum value is -22 at x = 3.

Method to apply

Find f'(x); solve f'(x) = 0 and list critical points in the domain; use first derivative sign change or second derivative test; calculate f(a) for each confirmed extremum; state whether it is local maximum or local minimum.

Diagram support

A curve sketch or derivative sign chart is useful to show rising-falling behaviour around a critical point.

How CBSE asks it

Appears as a derivative-test problem, assertion-based reasoning, graph interpretation, or an optimisation step inside a longer question.

Avoid common mistakes

Common confusion

Assuming every solution of f'(x) = 0 is automatically a maximum or minimum.

Common wrong answer

Using f''(a) > 0 as maximum and f''(a) < 0 as minimum, which reverses the second derivative test.

Exam tip

After finding a critical point, always apply a test and state both the point and the value if asked.

Quick check

If f'(a) = 0 and f''(a) = -5, what type of extremum may occur at x = a?

Since f''(a) < 0, f has a local maximum at x = a.

Answer writing and exam use

1-mark answer

A function has a local maximum at a point if its value is greater than or equal to nearby values, and a local minimum if its value is less than or equal to nearby values.

2-mark answer

A function has a local maximum at a point if its value is greater than or equal to nearby values, and a local minimum if its value is less than or equal to nearby values. First derivative test: f' changes from positive to negative at x = a gives local maximum; f' changes from negative to positive gives local minimum. Second derivative test: if f'(a) = 0 and f''(a) < 0, local maximum; if f''(a) > 0, local minimum; if f''(a) = 0, use another test. For f(x) = x^2 - 4x + 5, f'(x) = 2x - 4, so x = 2. Since f''(x) = 2 > 0, f has a local minimum at x = 2, and the minimum value is f(2) = 1.

3-mark answer

Possible points of maxima or minima occur at critical points. The first derivative test checks the sign change of f'(x) around the point. The second derivative test uses f''(a): if f'(a) = 0 and f''(a) < 0, there is a local maximum; if f''(a) > 0, there is a local minimum. If f''(a) = 0, the test is inconclusive. First derivative test: f' changes from positive to negative at x = a gives local maximum; f' changes from negative to positive gives local minimum. Second derivative test: if f'(a) = 0 and f''(a) < 0, local maximum; if f''(a) > 0, local minimum; if f''(a) = 0, use another test. Find local extrema of f(x) = x^3 - 3x^2 - 9x + 5. f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1), so critical points are x = -1 and x = 3. f''(x) = 6x - 6. At x = -1, f''(-1) = -12 < 0, so local maximum; f(-1) = -1 - 3 + 9 + 5 = 10. At x = 3, f''(3) = 12 > 0, so local minimum; f(3) = 27 - 27 - 27 + 5 = -22. Thus local maximum value is 10 at x = -1 and local minimum value is -22 at x = 3. Appears as a derivative-test problem, assertion-based reasoning, graph interpretation, or an optimisation step inside a longer question. Using f''(a) > 0 as maximum and f''(a) < 0 as minimum, which reverses the second derivative test.
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