1. Solve the inequalities and equations: I. Solve $$|2x + 3| \leq |2x + 1|$$ 1. Since both sides are absolute values, square both sides (valid since absolute values are non-negative): $$ (2x + 3)^2 \leq (2x + 1)^2 $$ 2. Expand both sides: $$ 4x^2 + 12x + 9 \leq 4x^2 + 4x + 1 $$ 3. Subtract $$4x^2$$ from both sides: $$ 12x + 9 \leq 4x + 1 $$ 4. Rearrange terms: $$ 12x - 4x \leq 1 - 9 $$ $$ 8x \leq -8 $$ 5. Divide both sides by 8: $$ x \leq -1 $$ II. Solve $$|x - 3| \leq 4$$ 1. The definition of absolute inequality: $$ -4 \leq x - 3 \leq 4 $$ 2. Add 3 to all sides: $$ -1 \leq x \leq 7 $$ III. Solve $$ \frac{x^2 - 3x - 54}{x^2 - 5x - 14} \geq 0 $$ 1. Factor numerator and denominator: Numerator: $$x^2 - 3x - 54 = (x - 9)(x + 6)$$ Denominator: $$x^2 - 5x - 14 = (x - 7)(x + 2)$$ 2. Critical points: $$x = -6, 7, -2, 9$$ 3. Determine where the rational expression is non-negative using sign analysis: Intervals: $$(-\infty, -6), (-6, -2), (-2, 7), (7, 9), (9, \infty)$$ Evaluate signs in each interval (taking into account denominator zeros are excluded): - $$(-\infty, -6)$$: numerator positive, denominator positive, expression positive. - $$(-6, -2)$$: numerator negative, denominator positive, expression negative. - $$(-2, 7)$$: numerator negative, denominator negative, expression positive. - $$ (7, 9) $$: numerator negative, denominator positive, expression negative. - $$ (9, \infty) $$: numerator positive, denominator positive, expression positive. 4. Include points where numerator is zero (zeros of numerator): $$x=-6,9$$ Denominator zeros $$x=-2,7$$ are excluded. 5. Solution: $$ (-\infty, -6] \cup (-2, 7) \cup [9, \infty) $$ IV. Solve $$ |3x - 18| > 6 $$ 1. Write inequality as: $$ |3(x-6)| > 6 $$ 2. Divide both sides by 3: $$ |x - 6| > 2 $$ 3. Using definition of absolute value inequality: $$ x - 6 < -2 \quad \text{or} \quad x - 6 > 2 $$ 4. Solve: $$ x < 4 \quad \text{or} \quad x > 8 $$ 2. Find domain and range of functions: I. $$f(x) = \sqrt{25 - x^2}$$ 1. Domain: $$25 - x^2 \geq 0 \Rightarrow x^2 \leq 25 \Rightarrow -5 \leq x \leq 5$$ 2. Range: Since $$f(x)$$ is a square root, minimum is 0 when $$x = \pm 5$$ and maximum when $$x=0$$: $$f(0) = \sqrt{25} = 5$$ Thus, range is $$[0, 5]$$ II. $$f(x) = 3 + \frac{1}{2x - 5}$$ 1. Domain: denominator cannot be zero $$2x - 5 \neq 0 \Rightarrow x \neq \frac{5}{2}$$ 2. Range: All real numbers except the horizontal asymptote. The horizontal asymptote is: $$ y = 3 $$ Thus range is $$(-\infty, 3) \cup (3, \infty)$$ III. $$f(x) = \sqrt{x^2 - 9}$$ 1. Domain: $$x^2 - 9 \geq 0 \Rightarrow x^2 \geq 9 \Rightarrow x \leq -3$$ or $$ x \geq 3$$ 2. Range: Since $$\sqrt{\cdot}$$ is non-negative, Minimum value is 0 (at $$x=\pm 3$$), range is $$[0, \infty)$$ IV. $$f(x) = -1 + \frac{1}{x^2 + 25}$$ 1. Domain: denominator $$x^2+25 > 0$$ always, so all real numbers 2. Range: Since $$x^2 \geq 0$$, minimum of denominator is 25 (at $$x=0$$). $$ \frac{1}{x^2 + 25} \leq \frac{1}{25} $$, and approaches 0 as $$|x| \to \infty$$ So: $$ f(x) = -1 + \frac{1}{x^2 + 25} \in \left(-1, -1 + \frac{1}{25}\right] = (-1, -0.96] $$ 3. Determine if graph is function and state domain + range: I. Given points (0,0), (1,2), (2,3) and smooth increasing curve - By vertical line test, passes for all x - Domain: $$[0,2]$$ - Range: $$[0,3]$$ II. Step function changing between y=1 and y=0 at integers 0,1,2 - Function since only one y per x - Domain: $$[0, \infty)$$ - Range: $$\{0, 1\}$$ 4. Check function parity: V. $$f(x) = \frac{x^3 + x - 1}{x^2 + 1}$$ - Test $$f(-x)$$: $$ \frac{-x^3 - x -1}{x^2 + 1} \neq f(x) \text{ and } \neq -f(x) $$ - So neither even nor odd VI. $$f(x) = \frac{x^3}{|x|}$$ - Simplify for $$x \neq 0$$: $$f(x) = x^2 \cdot \text{sign}(x)$$ - Test $$f(-x) = \frac{(-x)^3}{|-x|} = \frac{-x^3}{|x|} = -f(x)$$ - So function is odd VII. $$f(x) = \frac{1}{x^4 + x^2 + 5}$$ - Even powers only, so $$f(-x) = f(x)$$ - Function is even VIII. $$f(x) = x^2 - 6x + 9$$ - Test $$f(-x) = (-x)^2 - 6(-x) + 9 = x^2 + 6x + 9 \neq f(x)$$ - Not even or odd 5. Find intervals of increasing and decreasing: I. $$f(x) = 1 - 4x^5$$ 1. Derivative: $$f'(x) = -20x^4$$ 2. Since $$x^4 \geq 0$$, $$f'(x) \leq 0$$ always, derivative zero at $$x=0$$ 3. Function is decreasing everywhere except flat at zero II. $$f(x) = 4 - \frac{3}{\sqrt{x^2 + 1}}$$ 1. Derivative: $$f'(x) = -3 \cdot \left(-\frac{1}{2}\right) (x^2 + 1)^{-3/2} \cdot 2x = \frac{3x}{(x^2 + 1)^{3/2}}$$ 2. Sign of derivative depends on numerator: - For $$x > 0$$, $$f'(x) > 0$$ increasing - For $$x < 0$$, $$f'(x) < 0$$ decreasing - At $$x=0$$ derivative zero III. $$f(x) = \cos x$$ on $$[0, 3\pi]$$ 1. Derivative: $$f'(x) = -\sin x$$ 2. $$f'(x) > 0$$ when $$-\sin x > 0 \Rightarrow \sin x < 0$$ 3. Analyze on intervals: - Increasing where $$\sin x < 0$$, e.g., $$(\pi, 2\pi)$$ - Decreasing elsewhere IV. $$f(x) = \frac{1}{x^3 + 1}$$, domain $$x > 0$$ 1. Derivative: $$f'(x) = -\frac{3x^2}{(x^3 + 1)^2} < 0$$ 2. Always negative for $$x > 0$$, so decreasing 6. Determine if functions are bounded: V. $$f(x) = \frac{2}{1 + 5x^2}$$ - Denominator $$\geq 1$$ so $$f(x) \leq 2$$ - Minimum approaches 0 as $$|x| \to \infty$$ - Function is bounded VI. $$f(x) = \frac{1}{x+1} + \frac{1}{x^2 + 2}, x \in (0, 2]$$ - Denominator of both terms positive, expressions finite on domain - Each term bounded: - $$\frac{1}{x+1}$$ bounded on $$(0,2]$$ - $$\frac{1}{x^2 + 2}$$ bounded and positive - Function bounded on given domain