Exemplar Problem: Calculus - Multiple Solution Methods

📋 Problem Statement

Question: Find the minimum value of the function f(x) = x + 1/x for x > 0.

Given:

• Function: f(x) = x + 1/x
• Domain: x > 0
• Find: Minimum value of f(x)

🎯 Solution Method 1: Derivative Approach

Step 1: Find the First Derivative

Given: f(x) = x + 1/x

Differentiate with respect to x: f’(x) = d/dx (x + 1/x) f’(x) = 1 + d/dx (x⁻¹) f’(x) = 1 - x⁻² f’(x) = 1 - 1/x²

Step 2: Find Critical Points

Set f’(x) = 0 for critical points: 1 - 1/x² = 0 1 = 1/x² x² = 1 x = ±1

Considering domain x > 0: Critical point: x = 1

Step 3: Second Derivative Test

Find the second derivative: f’’(x) = d/dx (1 - 1/x²) f’’(x) = 0 - d/dx (x⁻²) f’’(x) = -(-2x⁻³) f’’(x) = 2/x³

Evaluate at x = 1: f’’(1) = 2/1³ = 2 > 0

Since f’’(1) > 0, x = 1 is a point of local minimum.

Step 4: Find Minimum Value

Evaluate f(x) at x = 1: f(1) = 1 + 1/1 = 1 + 1 = 2

Answer: The minimum value is 2, occurring at x = 1.

🎯 Solution Method 2: AM-GM Inequality Approach

Step 1: Apply AM-GM Inequality

Given: f(x) = x + 1/x, where x > 0

AM-GM Inequality: For positive numbers a and b, AM ≥ GM (a + b)/2 ≥ √(ab)

Let a = x and b = 1/x (both positive since x > 0): (x + 1/x)/2 ≥ √(x × 1/x) (x + 1/x)/2 ≥ √1 (x + 1/x)/2 ≥ 1 x + 1/x ≥ 2

Step 2: Find When Equality Occurs

AM = GM when a = b x = 1/x x² = 1 x = 1 (since x > 0)

Equality occurs at x = 1

Step 3: Minimum Value

From AM-GM inequality: x + 1/x ≥ 2 Minimum value: 2 Occurs at: x = 1

🎯 Solution Method 3: Completing the Square Approach

Step 1: Rewrite the Expression

Given: f(x) = x + 1/x

Let’s express in terms of √x: Let √x = t (since x > 0, √x is defined) Then x = t²

Substitute in f(x): f(x) = t² + 1/t²

Step 2: Complete the Square

Rearrange the expression: f(x) = t² + 1/t² f(x) = t² + 1/t² + 2 - 2 f(x) = (t + 1/t)² - 2

Step 3: Find Minimum

Since t + 1/t ≥ 2 (for t > 0): (t + 1/t)² ≥ 2² = 4

Therefore: f(x) = (t + 1/t)² - 2 ≥ 4 - 2 = 2

Minimum value: 2 Occurs when: t + 1/t = 2 Which means: t = 1 Therefore: √x = 1, so x = 1

🎯 Solution Method 4: Geometric Interpretation Approach

Step 1: Consider the Graph

Function: f(x) = x + 1/x Domain: x > 0

Analyze the behavior:

• As x → 0⁺: f(x) → +∞
• As x → +∞: f(x) → +∞
• The function must have a minimum somewhere in between

Step 2: Use Symmetry Properties

Consider the substitution x → 1/x: f(1/x) = (1/x) + x = x + 1/x = f(x)

The function is symmetric about x = 1 This suggests the minimum occurs at x = 1

Step 3: Verify by Testing Values

Test x = 1: f(1) = 1 + 1 = 2

Test x = 2: f(2) = 2 + 1/2 = 2.5 > 2

Test x = 1/2: f(1/2) = 1/2 + 2 = 2.5 > 2

Conclusion: Minimum occurs at x = 1 with value 2

🔍 Comparison of Methods

Method 1: Derivative Approach

• Pros: Rigorous, standard calculus method
• Cons: Requires differentiation skills
• Best for: Students comfortable with calculus

Method 2: AM-GM Inequality

• Pros: Elegant, no calculus required
• Cons: Limited to certain types of problems
• Best for: Inequality optimization problems

Method 3: Completing the Square

• Pros: Algebraic approach, visually intuitive
• Cons: Requires algebraic manipulation skills
• Best for: Quadratic-like expressions

Method 4: Geometric Interpretation

• Pros: Visual understanding, intuitive
• Cons: Less rigorous, depends on pattern recognition
• Best for: Developing mathematical intuition

💡 Mathematical Insights

Generalization:

For any positive number n: Function: f(x) = xⁿ + 1/xⁿ Minimum value: 2 Occurs at: x = 1

Extension:

Function: f(x) = ax + b/x (where a, b > 0) Derivative method: f’(x) = a - b/x² = 0 Critical point: x = √(b/a) Minimum value: 2√(ab)

Connection to Other Problems:

This type of expression appears in:

• Physics: Work-energy problems
• Economics: Cost minimization
• Engineering: Optimization problems

🎯 Applications and Extensions

Related Problems:

1. Find minimum of f(x) = x² + 1/x²
2. Find minimum of f(x) = 3x + 12/x for x > 0
3. Find minimum of f(x) = x + 4/x for x > 0

Challenge Problems:

1. Find minimum of f(x) = x + 1/x + 1/x² for x > 0
2. Find minimum of f(x) = x^3 + 1/x for x > 0
3. Find minimum of f(x) = x + 1/x + 2 for x > 0

Real-world Applications:

1. Economics: Cost minimization with economies of scale
2. Physics: Energy optimization in systems
3. Engineering: Resource allocation problems

📈 Graphical Analysis

Function Behavior:

f(x) = x + 1/x
Domain: x > 0

x    | f(x)
-----|-------
0.1  | 10.1
0.2  | 5.2
0.5  | 2.5
1    | 2    ← Minimum
2    | 2.5
5    | 5.2
10   | 10.1

Key Observations:

1. Symmetric about x = 1
2. Convex function for x > 0
3. Global minimum at x = 1
4. Minimum value: 2

🎯 Key Takeaways

Problem-Solving Strategies:

1. Multiple methods lead to same result
2. Choose method based on your strengths
3. Verify answers using alternative approaches
4. Look for patterns in mathematical expressions

Important Concepts:

1. Derivative test: First derivative zero, second derivative positive
2. AM-GM inequality: Arithmetic mean ≥ Geometric mean
3. Completing square: Transform to perfect square form
4. Symmetry: Mathematical properties simplify problems

General Tips:

• Always check domain restrictions
• Verify critical points are within domain
• Consider boundary behavior
• Use multiple methods for confidence

Remember: Optimization problems often have multiple solution methods. Understanding different approaches makes you a more versatile problem solver!

Happy Learning! 📐