🔄 Functions & Transformations
Functions are the foundation of precalculus. Understanding domains, ranges, inverses, and transformations is essential for AMC success.
🎯 Key Ideas
Function Definition: A function $f$ maps each input $x$ to exactly one output $f(x)$. The domain is all valid inputs; the range is all possible outputs.
Transformations: Basic transformations shift, scale, or reflect graphs:
- $f(x-h)$ shifts right by $h$ units
- $f(x)+k$ shifts up by $k$ units
- $af(x)$ scales vertically by factor $a$
- $f(bx)$ scales horizontally by factor $\frac{1}{b}$
📊 Domain and Range
Common Function Types
| Function | Domain | Range | Notes |
|---|---|---|---|
| $f(x) = x^2$ | $(-\infty, \infty)$ | $[0, \infty)$ | Non-negative |
| $f(x) = \sqrt{x}$ | $[0, \infty)$ | $[0, \infty)$ | Non-negative |
| $f(x) = \frac{1}{x}$ | $(-\infty, 0) \cup (0, \infty)$ | $(-\infty, 0) \cup (0, \infty)$ | Exclude 0 |
| $f(x) = \log_a x$ | $(0, \infty)$ | $(-\infty, \infty)$ | $a > 0, a \neq 1$ |
| $f(x) = a^x$ | $(-\infty, \infty)$ | $(0, \infty)$ | $a > 0, a \neq 1$ |
Pitfall: Domain Restrictions
- Square roots: radicand $\geq 0$
- Logarithms: argument $> 0$
- Fractions: denominator $\neq 0$
- Even roots: radicand $\geq 0$
🔄 Inverse Functions
Definition: $f^{-1}(x)$ is the function that “undoes” $f(x)$, so $f^{-1}(f(x)) = x$.
Finding Inverses:
- Replace $f(x)$ with $y$
- Swap $x$ and $y$
- Solve for $y$
- Replace $y$ with $f^{-1}(x)$
Example: Find inverse of $f(x) = 2x + 3$
- $y = 2x + 3$
- $x = 2y + 3$ (swap)
- $x - 3 = 2y$
- $y = \frac{x-3}{2}$
- $f^{-1}(x) = \frac{x-3}{2}$
Pitfall: Inverse vs Reciprocal
- $f^{-1}(x) \neq \frac{1}{f(x)}$ (inverse function vs reciprocal)
- Example: If $f(x) = 2x + 1$, then $f^{-1}(x) = \frac{x-1}{2}$, but $\frac{1}{f(x)} = \frac{1}{2x+1}$
🔄 Function Composition
Definition: $(f \circ g)(x) = f(g(x))$ (apply $g$ first, then $f$)
Order Matters: $(f \circ g)(x) \neq (g \circ f)(x)$ in general
Example: If $f(x) = x^2$ and $g(x) = x + 1$, then:
- $(f \circ g)(x) = f(x+1) = (x+1)^2$
- $(g \circ f)(x) = g(x^2) = x^2 + 1$
📈 Transformations
Vertical Transformations
- $f(x) + k$: shifts up $k$ units (down if $k < 0$)
- $af(x)$: scales vertically by factor $a$ (reflects if $a < 0$)
Horizontal Transformations
- $f(x-h)$: shifts right $h$ units (left if $h < 0$)
- $f(bx)$: scales horizontally by factor $\frac{1}{b}$ (reflects if $b < 0$)
Pro Move: Combined Transformations
For $af(b(x-h)) + k$:
- Apply horizontal shift: $f(x-h)$
- Apply horizontal scaling: $f(b(x-h))$
- Apply vertical scaling: $af(b(x-h))$
- Apply vertical shift: $af(b(x-h)) + k$
🎯 AMC-Style Worked Example
Problem: If $f(x) = x^2 + 2x$ and $g(x) = x + 1$, find $(f \circ g)(x)$ and its domain.
Solution:
- $(f \circ g)(x) = f(g(x)) = f(x+1)$
- $f(x+1) = (x+1)^2 + 2(x+1)$
- $= x^2 + 2x + 1 + 2x + 2$
- $= x^2 + 4x + 3$
Domain: Since $g(x) = x + 1$ has domain $(-\infty, \infty)$ and $f(x) = x^2 + 2x$ has domain $(-\infty, \infty)$, the composition has domain $(-\infty, \infty)$.
Answer: $(f \circ g)(x) = x^2 + 4x + 3$ with domain $(-\infty, \infty)$
🔍 Common Traps & Fixes
Trap: Confusing inverse and reciprocal
Fix: Remember $f^{-1}(x)$ undoes $f(x)$, while $\frac{1}{f(x)}$ is just the reciprocal.
Trap: Wrong order in composition
Fix: $(f \circ g)(x) = f(g(x))$ means apply $g$ first, then $f$.
Trap: Domain errors in composition
Fix: Check that outputs of inner function are valid inputs for outer function.
Trap: Forgetting horizontal scaling factor
Fix: $f(bx)$ scales horizontally by $\frac{1}{b}$, not $b$.
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