🎯 Functional Equation Templates — f(x+a)/f(-x) Invariances
Essential for AMC12 problems involving unknown functions and their properties.
🎯 Recognition Cues
- “Find $f(x)$” — Unknown function to be determined
- $f(x+a)$ — Function with shifted argument
- $f(-x)$ — Function with negated argument
- $f(xy) = f(x) + f(y)$ — Logarithmic-like properties
- $f(x+y) = f(x) + f(y)$ — Additive properties
📚 Template Solutions
Common Functional Equations
| Type | Equation | Solution | Example |
|---|---|---|---|
| Additive | $f(x+y) = f(x) + f(y)$ | $f(x) = cx$ | $f(2) = 2c$ |
| Multiplicative | $f(xy) = f(x) + f(y)$ | $f(x) = c\log x$ | $f(4) = c\log 4$ |
| Exponential | $f(x+y) = f(x)f(y)$ | $f(x) = a^x$ | $f(3) = a^3$ |
| Power | $f(xy) = f(x)f(y)$ | $f(x) = x^c$ | $f(4) = 4^c$ |
Substitution Strategies
| Pattern | Substitution | Purpose | Example |
|---|---|---|---|
| Zero | $x = 0$ | Find $f(0)$ | $f(x+y) = f(x) + f(y)$ → $f(0) = 0$ |
| Identity | $x = y$ | Find relationships | $f(x+y) = f(x) + f(y)$ → $f(2x) = 2f(x)$ |
| Negative | $x = -y$ | Exploit symmetry | $f(x+y) = f(x) + f(y)$ → $f(0) = f(x) + f(-x)$ |
| Reciprocal | $x = \frac{1}{y}$ | Find $f(1)$ | $f(xy) = f(x) + f(y)$ → $f(1) = 0$ |
🎯 Worked Examples
Example 1: Additive Functional Equation
Problem: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{R}$.
Solution:
- Substitute $x = 0$: $f(y) = f(0) + f(y)$ → $f(0) = 0$
- Substitute $x = -y$: $f(0) = f(x) + f(-x)$ → $f(-x) = -f(x)$ (odd function)
- Substitute $y = x$: $f(2x) = 2f(x)$
- By induction: $f(nx) = nf(x)$ for all integers $n$
- For rationals: $f(\frac{p}{q}x) = \frac{p}{q}f(x)$
- For reals: $f(x) = cx$ for some constant $c$ (assuming continuity)
- Answer: $f(x) = cx$ for some constant $c$
Example 2: Multiplicative Functional Equation
Problem: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ such that $f(xy) = f(x) + f(y)$ for all $x,y > 0$.
Solution:
- Substitute $x = y = 1$: $f(1) = f(1) + f(1)$ → $f(1) = 0$
- Substitute $y = \frac{1}{x}$: $f(1) = f(x) + f(\frac{1}{x})$ → $f(\frac{1}{x}) = -f(x)$
- Substitute $y = x$: $f(x^2) = 2f(x)$
- For rationals: $f(x^r) = rf(x)$ for rational $r$
- For reals: $f(x) = c\log x$ for some constant $c$ (assuming continuity)
- Answer: $f(x) = c\log x$ for some constant $c$
Example 3: Exponential Functional Equation
Problem: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$.
Solution:
- Substitute $x = y = 0$: $f(0) = f(0)^2$ → $f(0) = 0$ or $f(0) = 1$
- If $f(0) = 0$: $f(x) = f(x+0) = f(x)f(0) = 0$ for all $x$ (constant zero function)
- If $f(0) = 1$: $f(x) = f(x+0) = f(x)f(0) = f(x)$ (consistent)
- Substitute $y = -x$: $f(0) = f(x)f(-x)$ → $f(-x) = \frac{1}{f(x)}$ (assuming $f(x) \neq 0$)
- For rationals: $f(nx) = f(x)^n$ for integers $n$
- For reals: $f(x) = a^x$ for some constant $a > 0$ (assuming continuity)
- Answer: $f(x) = 0$ or $f(x) = a^x$ for some constant $a > 0$
⚠️ Common Pitfalls
Pitfall: Assuming continuity without justification
- Fix: Only assume continuity if explicitly stated or if it’s reasonable
- Example: $f(x+y) = f(x) + f(y)$ has discontinuous solutions without continuity
Pitfall: Forgetting to check all cases
- Fix: Consider all possible values of variables
- Example: $f(x^2) = f(x)^2$ has different cases for $x = 0, 1, -1$
Pitfall: Incorrect substitution
- Fix: Be careful with variable replacement
- Example: In $f(x+y) = f(x) + f(y)$, substituting $x = y$ gives $f(2x) = 2f(x)$, not $f(x^2) = 2f(x)$
🎯 AMC-Style Worked Example
Problem: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+1) = f(x) + 1$ and $f(x^2) = f(x)^2$ for all $x \in \mathbb{R}$.
Solution:
- From first equation: $f(x+1) = f(x) + 1$ for all $x$
- Substitute $x = 0$: $f(1) = f(0) + 1$
- From second equation: $f(x^2) = f(x)^2$ for all $x$
- Substitute $x = 0$: $f(0) = f(0)^2$ → $f(0) = 0$ or $f(0) = 1$
- If $f(0) = 0$: $f(1) = 1$
- If $f(0) = 1$: $f(1) = 2$
- Check $f(1) = 1$: $f(1^2) = f(1) = 1$ and $f(1)^2 = 1^2 = 1$ ✓
- Check $f(1) = 2$: $f(1^2) = f(1) = 2$ and $f(1)^2 = 2^2 = 4$ ✗
- So $f(0) = 0$ and $f(1) = 1$
- By first equation: $f(n) = n$ for all integers $n$
- By second equation: $f(x^2) = f(x)^2$ for all $x$
- For $x = \frac{1}{2}$: $f(\frac{1}{4}) = f(\frac{1}{2})^2$
- Answer: $f(x) = x$ for all $x$ (assuming continuity)
Key insight: Functional equations often require multiple substitutions and careful case analysis.
🔗 Related Patterns
- Functions — Functional equations involve unknown functions
- Symmetry — Look for patterns and relationships
- Substitution — Replace variables with specific values
- Continuity — Often needed for complete solutions
📝 Practice Checklist
- Master basic substitution techniques
- Practice symmetry recognition
- Learn common functional equation types
- Practice case analysis
- Understand continuity requirements
- Work on speed and accuracy
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