1. **Problem 12:** Find the equation of the tangent line to the curve $y = 2x \sin x$ at the point $\left(\frac{\pi}{2}, \pi\right)$. Step 1: Differentiate $y$ using the product rule: $$ y' = 2 \sin x + 2x \cos x $$ Step 2: Evaluate the derivative at $x = \frac{\pi}{2}$: $$ y'\left(\frac{\pi}{2}\right) = 2 \sin \frac{\pi}{2} + 2 \cdot \frac{\pi}{2} \cos \frac{\pi}{2} = 2 \cdot 1 + \pi \cdot 0 = 2 $$ Step 3: Use point-slope form for the tangent line: $$ y - \pi = 2 \left(x - \frac{\pi}{2}\right) $$ Simplify: $$ y = 2x - \pi + \pi = 2x $$ 2. **Problem 13:** Use implicit differentiation to find the tangent line of $\sin x + \cos y = 1$ at $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$. Step 1: Differentiate both sides with respect to $x$: $$ \cos x - \sin y \frac{dy}{dx} = 0 $$ Step 2: Solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{\cos x}{\sin y} $$ Step 3: Evaluate at $x = \frac{\pi}{2}$ and $y = \frac{\pi}{2}$: $$ \frac{dy}{dx} = \frac{\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}} = \frac{0}{1} = 0 $$ Step 4: Equation of tangent line using point-slope form: $$ y - \frac{\pi}{2} = 0 \cdot \left(x - \frac{\pi}{2}\right) $$ Simplify: $$ y = \frac{\pi}{2} $$ 3. **Problem 14:** Find the critical numbers of $f(x) = x^{\frac{3}{5}} (4 - x)$. Step 1: Differentiate using product rule: $$ f'(x) = \frac{3}{5} x^{-\frac{2}{5}} (4 - x) + x^{\frac{3}{5}} (-1) $$ Step 2: Simplify: $$ f'(x) = \frac{3}{5} x^{-\frac{2}{5}} (4 - x) - x^{\frac{3}{5}} $$ Rewrite $x^{\frac{3}{5}}$ as $x^{\frac{3}{5}} = x^{\frac{3}{5}} \cdot \frac{5}{5}$ for common denominator, or factor $x^{-\frac{2}{5}}$: $$ f'(x) = x^{-\frac{2}{5}} \left( \frac{3}{5} (4 - x) - x \right) $$ Step 3: Simplify inside parentheses: $$ \frac{3}{5} (4 - x) - x = \frac{12}{5} - \frac{3}{5} x - x = \frac{12}{5} - \frac{3}{5} x - \frac{5}{5} x = \frac{12}{5} - \frac{8}{5} x $$ Step 4: Set derivative equal to zero and find critical points: $$ x^{-\frac{2}{5}} \left( \frac{12}{5} - \frac{8}{5} x \right) = 0 $$ $x^{-\frac{2}{5}} = \frac{1}{x^{2/5}}$ undefined at $x=0$ so critical number. Solve: $$ \frac{12}{5} - \frac{8}{5} x = 0 \implies 12 - 8x = 0 \implies x = \frac{12}{8} = \frac{3}{2} $$ Critical numbers: $$ x = 0, \quad x = \frac{3}{2} $$ 4. **Problem 15:** Find horizontal asymptotes of $$ f(x) = \frac{\sqrt{x^{2} + 4}}{x - 2} $$ Step 1: Consider limits as $x \to \infty$ and $x \to -\infty$. For large $|x|$, $\sqrt{x^{2} + 4} \approx |x|$. Step 2: When $x \to \infty$, $$ f(x) \approx \frac{x}{x - 2} = \frac{x}{x(1 - 2/x)} = \frac{1}{1 - 0} = 1 $$ Step 3: When $x \to -\infty$, $$ f(x) \approx \frac{|x|}{x - 2} = \frac{-x}{x - 2} = \frac{-x}{x(1 - 2/x)} = \frac{-1}{1 - 0} = -1 $$ Step 4: Horizontal asymptotes are $$ y = 1 \quad \text{and} \quad y = -1 $$