Example 1 · level 1
Find the absolute minimum value of f(x) = x² + 2x + 4 on [−2, 0].
Show solution
- f′(x) = 2x + 2 = 0 → x = −1.
- f(−2) = 4, f(−1) = 3, f(0) = 4.
- Absolute minimum = 3 at x = −1.
Answer: 3
CLASS 12 MATHS · APPLICATIONS OF DERIVATIVES
Maxima and minima of a function
Find critical points where f′(x) = 0. On a closed interval, the absolute maximum and minimum are among f at the critical points and at the end points.
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Worked examples
Example 2 · level 2
Find the local minimum value of f(x) = 2x³ + 9x² + 6.
Show solution
- f′(x) = 6(x + 3)(x) = 0 → x = −3 or x = 0.
- f″(−3) = −18 < 0 → local maximum; f″(0) = 18 > 0 → local minimum.
- Local minimum = f(0) = 6.
Answer: 6
Example 3 · level 3
Find the absolute maximum value of f(x) = 2x³ − 15x² + 24x − 3 on the interval [0, 5].
Show solution
- f′(x) = 6(x − 1)(x − 4) = 0 → x = 1, 4.
- f(0) = −3, f(1) = 8, f(4) = −19, f(5) = −8.
- Absolute maximum = 8.
Answer: 8