1. Solve inequalities: I. Solve $|2x + 3| \leq |2x + 1|$: - Square both sides: $(2x+3)^2 \leq (2x+1)^2$. - Expand: $4x^2 + 12x + 9 \leq 4x^2 + 4x + 1$. - Subtract $4x^2$: $12x + 9 \leq 4x + 1$. - Simplify: $12x - 4x \leq 1 - 9 \Rightarrow 8x \leq -8$. - Divide: $x \leq -1$. - Check equality in absolute values region; solution is $x \leq -1$. II. Solve $|x - 3| \leq 4$: - Inequality means $-4 \leq x - 3 \leq 4$. - Add 3: $-1 \leq x \leq 7$. III. Solve $\frac{x^2 - 3x - 54}{x^2 - 5x - 14} \geq 0$: - Factor numerator: $x^2 - 3x -54 = (x - 9)(x + 6)$. - Factor denominator: $x^2 - 5x - 14 = (x - 7)(x + 2)$. - Critical points: $-6, 7, -2, 9$. - Test intervals: - $(-\infty, -6)$: Numerator $+$, Denominator $+$, quotient $+$. - $(-6, -2)$: Numerator $-$, Denominator $+$, quotient $-$. - $(-2, 7)$: Numerator $-$, Denominator $-$, quotient $+$. - $(7, 9)$: Numerator $-$, Denominator $+$, quotient $-$. - $(9, \infty)$: Numerator $+$, Denominator $+$, quotient $+$. - Exclude where denominator zero: $x \neq 7, -2$. - Solution: $(-\infty, -6] \cup (-2, 7) \cup [9, \infty)$. IV. Solve $|3x -18| > 6$: - Divide by 3: $|x -6| > 2$. - Which means $x -6 < -2$ or $x - 6 > 2$. - So $x < 4$ or $x > 8$. 2. Domain and range: I. $f(x) = \sqrt{25 - x^2}$: - Domain: $25 - x^2 \geq 0 \Rightarrow -5 \leq x \leq 5$. - Range: $f(x) \geq 0$; max at $x=0$ is 5, min 0. - Range: $[0,5]$. II. $f(x) = 3 + \frac{1}{2x - 5}$: - Denominator $\neq 0 \Rightarrow 2x - 5 \neq 0 \Rightarrow x \neq \frac{5}{2}$. - Domain: $\mathbb{R} \setminus \{2.5\}$. - Range: $\mathbb{R} \setminus \{3\}$ (since $f(x) \neq 3$). III. $f(x) = \sqrt{x^2 - 9}$: - Inside square root $\geq 0$: $x^2 \geq 9$. - Domain: $(-\infty, -3] \cup [3, \infty)$. - Range: $[0, \infty)$. IV. $f(x) = -1 + \frac{1}{x^2 + 25}$: - $x^2 + 25 > 0$ for all real x. - Domain: $\mathbb{R}$. - Range: $\frac{1}{x^2 + 25} \in (0, \frac{1}{25}]$, so $f(x) \in (-1, -1 + \frac{1}{25}] = (-1, -\frac{24}{25}]$. 3. Determine if curve is function, domain, range: I. Curve increasing from bottom left crossing $y=3$ near $x=4$: - Assuming vertical line passes once, function. - Domain: all $x$ shown in graph. - Range: all $y$ values from bottom to max. II. Step function with horizontal jumps at integers: - Vertical lines indicate multiple $y$ for some $x$? No, jumps occur only at points, not violating vertical line test. - It is a function. - Domain: integers and segments covered. - Range: values of steps. 4. Even/odd/neither: V. $f(x)=\frac{x^3 + x -1}{x^2+1}$ - Compute $f(-x) = \frac{-x^3 - x -1}{x^2 +1}$. - Not equal $f(x)$ nor $-f(x)$, so neither. VI. $f(x) = \frac{x^3}{|x|} = x^2 \frac{x}{|x|} = x^2 \operatorname{sgn}(x)$ - $f(-x) = (-x)^2 \operatorname{sgn}(-x) = x^2 (-1) = -f(x)$ - Odd function. VII. $f(x) = \frac{1}{x^4 + x^2 +5}$ - $f(-x) = f(x)$ - Even function. VIII. $f(x) = x^2 - 6x + 9$ - $f(-x) = x^2 + 6x + 9 \neq \pm f(x)$ - Neither even nor odd. 5. Intervals increasing/decreasing: I. $f(x) = 1 - 4x^5$ - $f'(x) = -20 x^4 \leq 0$ - Function decreasing everywhere except flat at $x=0$. II. $f(x) = 4 - \frac{3}{\sqrt{x^2 +1}}$ - $f'(x) = \frac{3x}{(x^2+1)^{3/2}}$. - Increasing where $x>0$, decreasing where $x<0$. III. $f(x) = \cos x$ on $[0, 3\pi]$ - Derivative $f'(x) = -\sin x$. - Increasing on $(\pi, 2\pi)$ where $\sin x < 0$. - Decreasing on $(0, \pi)$ and $(2\pi, 3\pi)$. IV. $f(x) = \frac{1}{x^3 +1}, x>0$ - Derivative: $f'(x) = -\frac{3x^2}{(x^3 +1)^2} <0$. - Decreasing on $(0, \infty)$. 6. Bounded or not: V. $f(x) = \frac{2}{1 + 5x^2}$ - Denominator $\geq 1$, function $\leq 2$ and $>0$. - Bounded. VI. $f(x) = \frac{1}{x+1} + \frac{1}{x^2 +2}, x \in (0, 2]$ - Near $x \to 0^+$, first term finite positive, second term finite. - At $x=2$, finite. - Function is bounded on domain. Final answer contained above.