Why do we care about peaks and valleys?
Imagine you’re planning a road trip and want to know the highest hill you’ll climb and the deepest dip you’ll cross. In maths, those "high points" and "low points" are called maxima and minima, and derivatives help us find them quickly.
💡 In Simple Words: A derivative tells us how fast something is changing, like the speed of a car. If the speed becomes zero and then starts to go the other way, we’ve hit a peak (maximum) or a trough (minimum).
Application of Derivatives: Maxima and Minima
In class 12 maths, the application of derivatives is a favorite topic for ISC exams because it turns abstract calculus into practical problem‑solving. The core idea is simple: wherever the slope of a curve (the derivative) is zero, the curve might be at a highest or lowest point.
Key Terms Explained
- Derivative: the instant rate of change of a function, like how fast water level rises in a pipe.
- Critical point: a point where the derivative is zero or undefined. Think of it as a traffic light where the car stops.
- Local maximum: the highest point in a small neighbourhood of the curve.
- Local minimum: the lowest point in a small neighbourhood.
- Second derivative test: checking the curvature (concavity) to decide if a critical point is a peak or a trough.
Step‑by‑Step Method (Flowchart)
Worked Example 1: A Simple Polynomial
Find the maximum and minimum of f(x)=x^3-3x^2+2.
- First derivative: f'(x)=3x^2-6x.
- Set to zero: 3x^2-6x=0 ⇒ 3x(x-2)=0 ⇒ x=0 or x=2. These are the critical points.
- Second derivative: f''(x)=6x-6.
- Test x=0: f''(0)=-6 (f(0)=2.
- Test x=2: f''(2)=6 (>0) → local minimum at x=2. Value: f(2)=2.
Notice the curve goes up, peaks at (0,2), dips down, then climbs again to a trough at (2,2). The numbers happen to be the same, but the shapes are opposite.
Worked Example 2: A Real‑World Cost Problem
A company’s cost (in thousands of rupees) to produce x thousand units is C(x)=5x^2-40x+150. Find the production level that gives minimum cost.
- Derivative: C'(x)=10x-40.
- Set to zero: 10x-40=0 ⇒ x=4 (thousand units).
- Second derivative: C''(x)=10 (>0) → the point is a minimum.
- Minimum cost: C(4)=5·16-40·4+150=80-160+150=70 (thousand rupees).
So producing 4,000 units saves the company money. This is exactly why exam questions love to wrap calculus in cost or profit stories.
Quick Revision Table
| Step | What to Do | Why? |
|---|---|---|
| 1 | Find f'(x) | Gets the slope of the curve. |
| 2 | Set f'(x)=0 | Finds points where slope stops – possible peaks or troughs. |
| 3 | Solve for x | These are the critical points. |
| 4 | Find f''(x) | Checks curvature: up‑ward means minimum, down‑ward means maximum. |
| 5 | Classify each point | Write down max/min values. |
Common Mistakes to Avoid
- Skipping the second derivative test and assuming every critical point is a max or min.
- Forgetting to check endpoints when the domain is closed (e.g., x∈[a,b]).
- Mixing up “local” (nearby) with “global” (overall) extremes.
📝 Likely Exam Questions
- Question: Find the maximum value of f(x)=2x^3-9x^2+12x+1 for x≥0.
Answer: f'(x)=6x^2-18x+12=0 ⇒ x=1 or x=2. f''(1)=-6 ( - Question: A rectangular garden has a fixed perimeter of 100 m. Using calculus, find the dimensions that give the maximum area.
Answer: Let length = x, width = 50‑x. Area A = x(50‑x)=50x‑x^2. A' = 50‑2x=0 ⇒ x=25 m, width=25 m. Max area = 625 m². - Question: The profit function is P(x)= -3x^2+30x‑20. Determine the production level for maximum profit and the profit amount.
Answer: P'(x)= -6x+30=0 ⇒ x=5. P''(x)= -6 ( - Question: Explain why a point where f'(x)=0 but f''(x)=0 needs further testing.
Answer: The second‑derivative test is inconclusive; we must use higher‑order derivatives or examine sign changes of f'(x) around the point.