Undefined Functions: Complete SAT Guide
Master every scenario where a function becomes undefined — a critical SAT Math skill
1. Introduction: Defined vs. Undefined
A function $f(x)$ is defined at $x = a$ if $f(a)$ produces a real number output. A function is undefined at $x = a$ if $f(a)$ does not produce a real number.
✅ Defined
$f(x) = \frac{1}{x-3}$ at $x = 5$:
$f(5) = \frac{1}{5-3} = \frac{1}{2}$ — a real number ✓
❌ Undefined
$f(x) = \frac{1}{x-3}$ at $x = 3$:
$f(3) = \frac{1}{3-3} = \frac{1}{0}$ — division by zero ✗
Why Do Functions Become Undefined?
Certain mathematical operations have built-in restrictions:
| Operation | Undefined When | Reason |
|---|---|---|
| $\frac{a}{b}$ | $b = 0$ | Division by zero is impossible |
| $\sqrt{a}$ (even root) | $a < 0$ | No real square root of a negative |
| $\frac{1}{\sqrt{a}}$ | $a \leq 0$ | Negative under radical OR zero in denominator |
| $\log(a)$ or $\ln(a)$ | $a \leq 0$ | Log of zero or negative is undefined |
2. Division by Zero
Why is $\frac{a}{0}$ Undefined?
If $\frac{5}{0} = ?$, then $? \times 0 = 5$. But any number times zero equals zero, never 5. So no answer exists. It's not infinity, not zero — it's simply undefined.
🔑 Key Rule: Finding Undefined Values in Fractions
Set the denominator equal to zero → Solve for $x$ → Those $x$-values make the function undefined.
Example Walkthrough
Example A: $f(x) = \frac{1}{x - 3}$
Step 1: Set denominator = 0:
$x - 3 = 0$
Step 2: Solve:
$x = 3$
Result: $f(x)$ is undefined at $x = 3$
Example B: $f(x) = \frac{5}{2x + 8}$
Step 1: Set denominator = 0:
$2x + 8 = 0$
Step 2: Solve:
$2x = -8 \Rightarrow x = -4$
Result: Undefined at $x = -4$
Factored Denominators
$f(x) = \frac{x+1}{x^2 - 9}$
Factor: $x^2 - 9 = (x-3)(x+3)$
Set each factor = 0:
$x - 3 = 0 \Rightarrow x = 3$
$x + 3 = 0 \Rightarrow x = -3$
Undefined at $x = 3$ and $x = -3$
$f(x) = \frac{x-2}{x^2 - 5x + 6}$
Factor: $x^2 - 5x + 6 = (x-2)(x-3)$
Set each factor = 0:
$x = 2$ or $x = 3$
Undefined at $x = 2$ AND $x = 3$
⚠️ Even though $x = 2$ cancels with the numerator, the original function is still undefined there!
⚠️ Critical SAT Trap: The Numerator Doesn't Matter!
Consider $f(x) = \frac{x-2}{x-2}$. At $x = 2$: $f(2) = \frac{0}{0}$, which is indeterminate (a special type of undefined). Even though this "simplifies" to 1, the original function is undefined at $x = 2$. The graph has a hole at $x = 2$.