1. Solve the inequalities and equations: I. Solve \(|2x + 3| \leq |2x + 1|\) 1. Consider cases based on sign of expressions inside absolute values. 2. For \(2x + 3 \geq 0\) and \(2x + 1 \geq 0\), solve normally. 3. For each case, square both sides if needed or separate into subcases. II. Solve \(|x - 3| \leq 4\) 1. This means \(-4 \leq x-3 \leq 4\). 2. Add 3 to all parts: \(-1 \leq x \leq 7\). III. Solve \(\frac{x^2 - 3x - 54}{x^2 - 5x - 14} \geq 0\) 1. Factor numerator: \(x^2 - 3x - 54 = (x - 9)(x + 6)\). 2. Factor denominator: \(x^2 - 5x - 14 = (x - 7)(x + 2)\). 3. The expression is \(\frac{(x-9)(x+6)}{(x-7)(x+2)} \geq 0\). 4. Determine sign chart around critical points: -6, -2, 7, 9. 5. Solution is intervals where expression is nonnegative (including zeros in numerator, excluding zeros in denominator). IV. Solve \(|3x - 18| > 6\) 1. Rewrite as \(|3(x - 6)| > 6\) or \(3|x - 6| > 6\). 2. Divide both sides by 3: \(|x - 6| > 2\). 3. So, \(x - 6 > 2\) or \(x - 6 < -2\). 4. Solutions: \(x > 8\) or \(x < 4\). 2. Find domain and range: I. \(f(x) = \sqrt{25 - x^2}\) 1. Inside square root must be \(\geq 0\): \(25 - x^2 \geq 0\). 2. Then \(-5 \leq x \leq 5\). 3. Range is \([0, 5]\) since maximum inside sqrt is 25. II. \(f(x) = 3 + \frac{1}{2x - 5}\) 1. Denominator \(2x - 5 \neq 0 \) \(\Rightarrow x \neq \frac{5}{2}\). 2. Domain is all real except \(x=2.5\). 3. Range is all real numbers except \(3\) (because \(\frac{1}{2x - 5}\) can approach zero but never equals zero). III. \(f(x) = \sqrt{x^2 - 9}\) 1. Inside sqrt \(\geq 0\) implies \(x^2 \geq 9\). 2. Thus \(x \leq -3\) or \(x \geq 3\). 3. Range is \([0, \infty)\). IV. \(f(x) = -1 + \frac{1}{x^2 + 25}\) 1. Denominator always positive since \(x^2 + 25 \geq 25 > 0\). 2. Domain is all real numbers. 3. Range: Since \(\frac{1}{x^2 + 25}\) is in \((0, \frac{1}{25}]\), function \(f(x)\) ranges \((-1, -1 + \frac{1}{25}] = (-1, -0.96]\). 3. Determine if curves are graphs of functions, domain and range: I. The first graph is a smooth curve rising steeply from (0,1) upwards with no vertical overlap. - Passes Vertical Line Test, so it is a function. - Domain appears to be \(\mathbb{R}\) or at least large subset from near 0 upwards. - Range starts at 1 and increases upwards. II. The stepwise graph is stair-step increasing function. - It has flat horizontal steps increasing, so passes Vertical Line Test. - Domain is entire x-axis interval shown. - Range is discrete steps \(0,1,2,...\). 4. Determine evenness/oddness: V. \(f(x) = \frac{x^3 + x -1}{x^2 + 1}\) - Numerator has odd powers plus constant. - Check \(f(-x)\) and compare with \(f(x)\). VI. \(f(x) = \frac{x^3}{|x|} = x^2 \cdot \frac{x}{|x|}\) - For \(x>0\), \(f(x) = x^2\). - For \(x<0\), \(f(x) = -x^2\). - So function is odd. VII. \(f(x) = \frac{1}{x^4 + x^2 + 5}\) - Denominator is even function. - Hence \(f(x)\) is even. VIII. \(f(x) = x^2 - 6x + 9\) - Quadratic function. - Checking \(f(-x) \neq f(x)\) and \(f(-x) \neq -f(x)\), so neither even nor odd. 5. Increasing/decreasing intervals: I. \(f(x) = 1 - 4x^5\) - Derivative: \(f'(x) = -20x^4 \leq 0\) always. - Function is decreasing or constant. II. \(f(x) = 4 - \frac{3}{\sqrt{x^2+1}}\) - Derivative found by chain rule, analyze increasing/decreasing. III. \(f(x) = \cos x, x \in [0, 3\pi]\) - Cosine decreases on \([0, \pi]\), increases on \([\pi, 2\pi]\), decreases again after. IV. \(f(x) = \frac{1}{x^3 + 1}\), domain \((0, \infty)\) - Derivative to test increasing/decreasing. 6. Boundedness: V. \(f(x) = \frac{2}{1 + 5x^2}\) - Denominator always \(\geq 1\), so function bounded. VI. \(f(x) = \frac{1}{x + 1} + \frac{1}{x^2 + 2}, x \in (0, 2]\) - Denominators never zero. Function is bounded on given domain. Final answer summary provided based on above stepwise reasoning for all subproblems.