Interval Notation and Inequalities

The denominator is $0$ at $x = 2$.

• Correct: Every real number except $2$:
• Every real number except $-2$:
• $x > 2$:
• Every real number:

$2(-3) + 3 = -3$.

• Correct: $-3$:
• $9$:
• $-9$:
• $3$:

A bracket means the endpoint is included ($\le$ or $\ge$); a parenthesis means it is not ($<$ or $>$).

• $(-1, 4]$: The brackets are swapped: $-1$ is included (bracket) and $4$ is not (parenthesis).
• Correct: $[-1, 4)$: Right: $\le$ includes $-1$, so a bracket; $<$ leaves out $4$, so a parenthesis.
• $[-1, 4]$: This includes $4$, but $x < 4$ leaves $4$ out.
• $(-1, 4)$: This leaves out $-1$, but $x \ge -1$ includes it.

Subtract $3$: $-2x > 4$. Divide by $-2$ and flip: $x < -2$, which is $(-\infty, -2)$. Check $x = -3$: $3 + 6 = 9 > 7$.

• $(-2, \infty)$: This forgets to flip the inequality when dividing by $-2$. Check $x = 0$: $3 > 7$ is false.
• $(-\infty, 2)$: $-2x > 4$ gives $x < -2$, not $x < 2$: $4$ divided by $-2$ is $-2$. Dividing by a negative flips the inequality, and the bound comes out negative.
• Correct: $(-\infty, -2)$: Right: $-2x > 4$, and dividing by $-2$ flips the sign: $x < -2$.
• $(-\infty, -2]$: $x = -2$ gives $3 + 4 = 7$, and $7 > 7$ is false, so $-2$ is not included.

Do the same step to all three parts: subtract $3$ to get $-4 \le 2x < 6$, then divide by $2$ to get $-2 \le x < 3$.

• Correct: $[-2, 3)$: Right: $-1 \le 2x + 3 < 9$ becomes $-4 \le 2x < 6$, so $-2 \le x < 3$.
• $[-1, 9)$: These are the outputs. The question asks for the inputs: solve for $x$.
• $[-2, 3]$: $x = 3$ gives $f(3) = 9$, and $9 < 9$ is false, so $3$ is left out.
• $[-4, 6)$: This stops after subtracting $3$; divide every part by $2$ as well.

The inside of a square root cannot be negative: $6 - 2x \ge 0$, so $-2x \ge -6$ and $x \le 3$ (flip when dividing by $-2$).

• $[3, \infty)$: Dividing $-2x \ge -6$ by $-2$ flips the sign: $x \le 3$, not $x \ge 3$.
• $(-\infty, 3)$: $x = 3$ gives $\sqrt{0} = 0$, a real number, so $3$ is included.
• $(-\infty, 6]$: Solve $6 - 2x \ge 0$ for $x$: the $6$ is divided by $2$ as well.
• Correct: $(-\infty, 3]$: Right: $6 - 2x \ge 0$ gives $x \le 3$.

Only $x = 5$ is left out. Skipping one number splits the line into two intervals, joined with the union sign $\cup$.

• Correct: $(-\infty, 5) \cup (5, \infty)$: Right: every real number except $5$, written as two intervals joined by $\cup$.
• $(-\infty, 5] \cup [5, \infty)$: Brackets at $5$ would include it, but $h(5)$ divides by $0$.
• $(5, \infty)$: Numbers below $5$ are fine: $h(0) = -\frac{4}{5}$.
• $(-\infty, \infty)$: At $x = 5$ the denominator is $0$, so $5$ is not in the domain.

$(-2, 5]$ means $-2 < x \le 5$: $-2$ is out, $5$ is in, and so is everything between.

• $-2$: The parenthesis at $-2$ leaves it out.
• Correct: $0$: In: $-2 < 0 \le 5$.
• Correct: $5$: In: the bracket at $5$ includes it.
• $5.5$: Out: $5.5$ is past the right end, $5$.