1. **Sketch and analyze** $f(x)=\frac{x^2 - 81}{x + 9}$. Simplify numerator: $x^2 - 81 = (x - 9)(x + 9)$. The function simplifies to $f(x) = x - 9$ for $x \neq -9$ (since denominator zero at $x=-9$). Domain: All real $x$ except $x=-9$. Range: All real numbers. Vertical asymptote at $x=-9$ (division by zero). Root at $x=9$ ($f(9)=0$). 2. **Piecewise function** $f(x) = \begin{cases} 1 & x \leq -1 \\ -x & -1 < x < 0 \\ 1 & x=0 \\ -x & 0 < x < 1 \\ 1 & x \geq 1 \end{cases}$ Domain: All real numbers. Range: From $-1$ to $1$, since $f(x) = -x$ varies between these values on intervals $(-1,0)$ and $(0,1)$, and 1 elsewhere. 3. **Sum of functions** $f+g$: $f(x) = \sqrt{x+2}$, $g(x) = x^2 -4$. $(f+g)(x) = \sqrt{x+2} + x^2 - 4$ Domain: $x+2 \geq 0 \Rightarrow x \geq -2$, no restriction from $g$. Range: Minimum depends on values, but since $x^2-4$ is unbounded, range is all real. 4. **Difference** $(f-g)(x) = \sqrt{x+2} - (x^2 -4) = \sqrt{x+2} - x^2 + 4$ Domain: $x \geq -2$ 5. **Product** $(fg)(x) = \sqrt{x+2} (x^2 -4)$ Domain: $x \geq -2$ 6. **Quotient** $(f/g)(x) = \frac{\sqrt{x+2}}{x^2 - 4}$ Domain: $x \geq -2$ and $x^2 -4 \neq 0 \Rightarrow x \neq \pm 2$. 7. **Composition** $f \circ g (x) = f(g(x)) = \sqrt{g(x)+ 2} = \sqrt{x^2 -4 + 2} = \sqrt{x^2 -2}$ Domain: $x^2 - 2 \geq 0 \Rightarrow |x| \geq \sqrt{2}$ 8. **Composition** $g \circ f (x) = g(f(x)) = (\sqrt{x+2})^2 -4 = x + 2 -4 = x - 2$ Domain: $x \geq -2$ Range: All real. 9. **Limit** $\lim_{x\to 1} \frac{1 - \sqrt{x}}{1 - x}$ Rationalize numerator: $$\frac{1 - \sqrt{x}}{1 - x} \cdot \frac{1 + \sqrt{x}}{1 + \sqrt{x}} = \frac{1 - x}{(1 - x)(1 + \sqrt{x})} = \frac{1}{1 + \sqrt{1}} = \frac{1}{2}$$ 10. **Limit** $\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} (2x + h) = 2x$ 11. **Limit** $\lim_{x \to \infty} \frac{x^4 + x^3}{12x^3 + 128} = \lim_{x \to \infty} \frac{x^3(x + 1)}{x^3(12 + 128/x^3)} = \lim_{x \to \infty} \frac{x+1}{12 + 0} = \infty$ 12. **Limit** $\lim_{x \to -4} \frac{2x}{16 - x^2} = \frac{2(-4)}{16 - 16} = \frac{-8}{0}$, limit does not exist, vertical asymptote. 13. **Limit** $\lim_{x \to 2^-} \frac{x-1}{x^2 - 4} = \lim_{x \to 2^-} \frac{x-1}{(x-2)(x+2)}$ Denominator approaches zero from negative side, numerator approaches 1. So, limit tends to $-\infty$. 14. **Piecewise** $f(x) = \begin{cases} x^2 - 1 & x < 3 \\ x + 5 & x \geq 3 \end{cases}$ Check continuity at $x=3$: Limit from left: $3^2 -1 = 8$ Value at 3: $3 + 5 = 8$ Continuous at $x=3$. 15. **Piecewise** $f(x) = \begin{cases} \frac{x-4}{\sqrt{16 - x^2}} & x < -4 \\ 4 - x & -4 \leq x \leq 4 \\ 4 & x > 4 \end{cases}$ Domain: For root real, $16 - x^2 \geq 0 \Rightarrow -4 \leq x \leq 4$; function undefined outside this except indicated. Check for continuity at $x=\pm 4$ by evaluating limits and function values. **Derivative problems:** 1. $y = \frac{3x - 1}{2x + 5}$ Using quotient rule: $$y' = \frac{(3)(2x + 5) - (3x - 1)(2)}{(2x + 5)^2} = \frac{6x + 15 - 6x + 2}{(2x + 5)^2} = \frac{17}{(2x+5)^2}$$ 2. $y = \sqrt{a^2 + x^2} = (a^2 + x^2)^{1/2}$ $$y' = \frac{1}{2} (a^2 + x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{a^2 + x^2}}$$ 3. $h(t) = (t^4 - 3t^2 + 5)(3t^4 - t^{-2})$ Use product rule: $$h' = (4t^3 - 6t)(3t^4 - t^{-2}) + (t^4 - 3t^2 + 5)(12t^3 + 2t^{-3})$$ 4. $$y = \frac{(3x - 2)^{-2}}{(2x - 6)^{-2}} = (3x - 2)^{-2} \cdot (2x - 6)^2$$ Differentiate using product rule and chain rule. 5. $$x = \frac{t^2 + 2}{(t + 4)^2}, \quad y = \frac{t}{t+4}$$ Differentiate $x(t)$ and $y(t)$ with quotient and chain rules accordingly. 6. $$y = u^2 + 9, \quad u = 4x^3 + 9$$ By chain rule: $$y' = 2u \cdot u' = 2(4x^3 + 9)(12x^2) = 24x^2 (4x^3 + 9)$$ 7. $$f(x) = -\frac{4}{(t + 2)^2} = -4(t+2)^{-2}$$ $$f' = -4 \cdot (-2)(t+2)^{-3} = \frac{8}{(t+2)^3}$$ Second derivative: $$f'' = 8 \cdot (-3)(t+2)^{-4} = -\frac{24}{(t+2)^4}$$ Third derivative: $$f''' = -24 \cdot (-4)(t+2)^{-5} = \frac{96}{(t+2)^5}$$ 8. $$y = \frac{1}{\sqrt{x}} + \sqrt{x} = x^{-1/2} + x^{1/2}$$ $$y' = -\frac{1}{2}x^{-3/2} + \frac{1}{2}x^{-1/2} = \frac{1}{2} \left(x^{-1/2} - x^{-3/2}\right)$$ 9. Implicit differentiation: $$\sqrt{x} + y + xy = 21$$ Differentiate both sides: $$\frac{1}{2\sqrt{x}} + y' + y + x y' = 0$$ Group $y'$ terms: $$y'(1 + x) = - y - \frac{1}{2\sqrt{x}}$$ $$y' = \frac{- y - \frac{1}{2\sqrt{x}}}{1 + x}$$ 10. $$x^2 y^2 = x^2 + y^2$$ Differentiate both sides: $$2x y^2 + x^2 (2y y') = 2x + 2y y'$$ Group $y'$ terms: $$2x^2 y y' - 2y y' = 2x - 2x y^2$$ $$y'(2x^2 y - 2 y) = 2x (1 - y^2)$$ $$y' = \frac{2x (1 - y^2)}{2y(x^2 - 1)} = \frac{x (1 - y^2)}{y (x^2 - 1)}$$