EXAM ONE · MULTIPLE CHOICE

Domain & Interval Notation

Two rules. One notation system. This is rust, not confusion — you already know why the rules exist, the job is making them fire fast under pressure.

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The rules — all of them

What Actually Restricts a Domain

Denominator? Set it ≠ 0. ("You can always not have pizza, but the pizza has to exist" — a zero denominator means the fraction doesn't exist at all.)

Radical in the numerator (or on its own)? Set the radicand ≥ 0. Zero is fine — you can take the square root of 0.

Radical in the denominator? Set the radicand > 0 — strict, not ≥. Zero there would make the denominator zero, and that's a separate, harder rule than the plain radical rule.

A radical rule and a denominator rule are independent — a negative number breaks a square root regardless of whether it's on top or bottom; a zero only breaks a denominator, never a numerator.

Sign-flip alert: when you multiply or divide an inequality by a negative number, the direction flips. "$-x > -1$" becomes "$x < 1$" — not "$x > 1$." This is the single most common place this rust shows up, in domain problems and everywhere else.

Quick check — before you start

One Question, 60 Seconds

Work this on paper first. Then check — this tells you whether to breeze through the practice set or slow down on the worked examples first.

$$f(x) = \dfrac{1}{x-3}$$

Find the domain, in interval notation.

A) $(-\infty, 3) \cup (3, \infty)$

B) $(-\infty, 3] \cup [3, \infty)$

C) $[-\infty, 3) \cup (3, \infty)$

D) $(-\infty, 3)$

Check it ›

A is correct. Denominator ≠ 0 → $x \neq 3$. Parenthesis on both sides of 3 because 3 is excluded, union covers everything else.

If you picked B — you treated $x=3$ as included instead of excluded. The excluded value never gets a bracket. Go re-read "the pizza has to exist" above, then hit the first worked example before the practice set.

If you picked C — infinity never gets a bracket, ever. You can't "reach" infinity, so it's always a parenthesis regardless of which rule excluded the other side.

If you picked D — you found the restriction but dropped the union. $x=3$ is bad; everything else is fine, and interval notation has to say both halves.

Got A cold? Skip straight to the practice set below.

Worked Example 1 — Radical on Top

$$f(x) = \frac{\sqrt{x+7}}{(x+3)(x-2)}$$

1 · Numerator has a radical (on its own, not in a denominator) → non-strict: $x+7 \geq 0 \Rightarrow x \geq -7$.

2 · Denominator → set each factor $\neq 0$: $x+3=0 \Rightarrow x=-3$, and $x-2=0 \Rightarrow x=2$.

3 · Combine. Everything from $-7$ onward is allowed, except $x=-3$ and $x=2$ punch holes in it.

4 · Write it in order. $-7$ is the leftmost boundary (bracket — it's allowed), then $-3$ (excluded), then $2$ (excluded), then on to infinity.

$$[-7, -3) \cup (-3, 2) \cup (2, \infty)$$

The trap: $-7$ gets a bracket (radical rule, non-strict) but $-3$ and $2$ get parentheses (denominator rule, always strict). Two different rules can sit right next to each other in the same answer with different bracket types — don't let one rule's bracket style bleed into the other.

Worked Example 2 — Radical on the Bottom (Strict)

$$f(x) = \frac{x+5}{\sqrt{x+12}}$$

1 · Numerator is just $x+5$ — no radical, no denominator. It can be anything, including negative or zero. "The numerator can always not have pizza."

2 · Denominator has a radical. Two rules stack here: it's a radical (needs the radicand non-negative) and it's a denominator (needs to not be zero). Combined, that means strict: $x+12 > 0$, not $\geq 0$.

3 · Why strict? If $x+12=0$, the square root itself is $0$ — a legal radical — but then the whole denominator is $0$, which isn't legal. The radical rule alone would allow it; the denominator rule kills it. When both rules touch the same spot, the stricter one wins.

4 · Solve. $x+12 > 0 \Rightarrow x > -12$.

$$(-12, \infty)$$

The trap: a radical by itself in a numerator gets $\geq$ and a bracket. The same radical moved into a denominator gets $>$ and a parenthesis — same-looking symbol, different rule, because the denominator rule is standing behind it. Ask "is this radical also a denominator?" before you write the inequality sign.

Now you

Faded — Fill the Blanks

Same machine as the worked examples. Don't peek — fill each blank, then open the answer.

$$f(x) = \frac{\sqrt{x-1}}{\sqrt{9-x}}$$

1 · Numerator radical, non-strict: $x - 1 \geq 0 \Rightarrow x \geq \underline{\quad}$

2 · Denominator radical, strict (it's also a denominator): $9-x > 0 \Rightarrow x < \underline{\quad}$

3 · Careful — solving $9-x>0$ means moving $x$ across the sign. Did you flip the direction correctly?

4 · Combine into interval notation: $\underline{\qquad}$

Check it ›

$x \geq 1$ (bracket, numerator radical, non-strict). $9-x>0 \Rightarrow -x>-9 \Rightarrow x<9$ (parenthesis — and notice the inequality flipped when we multiplied by $-1$). Combined: $[1, 9)$.

Practice Set (6) — Multiple Choice, Sheet Face-Down First

Work every problem on paper before you look at the options. The real exam is multiple choice — this is that format, with the wrong answers built from your actual error patterns.

1 · Rational, single linear denominator.

$f(x) = \dfrac{2x+1}{x-4}$. Domain?

A) $(-\infty,4)\cup(4,\infty)$ B) $(-\infty,4]\cup[4,\infty)$ C) $(-\infty,4)$ D) $[4,\infty)$

Check it ›

A. $x \neq 4$, parenthesis both sides (denominator rule is always strict), union covers the rest. B repeats the "included boundary" bracket error. C and D each drop half the domain.

2 · Radical, linear radicand (paired with #1 — same boundary, different rule).

$f(x) = \sqrt{x-4}$. Domain?

A) $(4,\infty)$ B) $[4,\infty)$ C) $(-\infty,4]$ D) $(-\infty,4)$

Check it ›

B. Radical alone, non-strict → $x \geq 4$, bracket. Notice #1 and #2 restrict at the same number (4) but need opposite bracket types — that's the denominator-rule-vs-radical-rule contrast, not a coincidence.

3 · Rational, factorable quadratic denominator.

$f(x) = \dfrac{x}{x^2-x-6}$. Domain?

A) $(-\infty,\infty)$ B) $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$ C) $(-\infty,-2)\cup(3,\infty)$ D) $(-\infty,3)\cup(3,\infty)$

Check it ›

B. Factor first: $(x-3)(x+2)$, zeros at $3$ and $-2$. Two exclusions means three pieces, unioned. C drops the middle interval; D only caught one zero.

4 · Radical restricted from above — bracket-direction check.

$f(x) = \sqrt{8-x}$. Domain?

A) $(-\infty,8]$ B) $[8,\infty)$ C) $(-\infty,8)$ D) $(8,\infty)$

Check it ›

A. $8-x \geq 0 \Rightarrow -x \geq -8 \Rightarrow x \leq 8$ — the inequality flips because you divided by $-1$. B is the classic "bracket always goes on the left" muscle-memory error; here the allowed side is actually going toward negative infinity.

5 · Distractor check — is the denominator actually ever zero?

$f(x) = \dfrac{x-1}{x^2+4}$. Domain?

A) $(-\infty,\infty)$ B) $(-\infty,-4)\cup(-4,\infty)$ C) $x \neq \pm 2$ D) $[1,\infty)$

Check it ›

A. $x^2+4=0$ has no real solution — a squared term plus a positive number is never zero. Nothing to exclude. B and C assume a denominator restriction exists without actually testing it; D confuses this with a radical-in-the-numerator problem.

6 · Fusion — radical over linear, both rules at once.

$f(x) = \dfrac{\sqrt{x+2}}{x-5}$. Domain?

A) $[-2,5)\cup(5,\infty)$ B) $[-2,\infty)$ C) $(-2,5)\cup(5,\infty)$ D) $[-2,5]$

Check it ›

A. Radical on top (non-strict): $x \geq -2$, bracket. Denominator (strict, always): $x \neq 5$, parenthesis on both sides of the hole. Combine: $[-2,5)\cup(5,\infty)$. B forgot the denominator rule entirely; D put a bracket at 5, which a denominator restriction never gets.

Before You Leave — Add This to Your Sheet

Your own line from session, and it's exactly right: "Close dot = bracket. Open = parenthesis. Infinity = always parenthesis." Add the rule that goes with it: denominator restrictions are always open/parenthesis; a radical's own restriction is closed/bracket unless that same radical is also sitting in a denominator.

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