Let's go through the solutions starting from Question 3 as requested.

Question 3

The question requires finding the tangent line $t_2$ to the curve $y = \frac{2}{3} x - \frac{7}{2} + \frac{1}{2x}$ at a point where $x = 3$ and where $t_2$ is parallel to $t_1$, the tangent at $x = 3$. We need to find the coordinates of this point and the equation of the tangent $t_2$.

1. Find the derivative $\frac{dy}{dx}$ to get the slope of the tangent lines: $y = \frac{2}{3}x - \frac{7}{2} + \frac{1}{2x}$ Differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{2}{3} - \frac{1}{2x^2}$

2. Find the slope of $t_1$ at $x = 3$: Substitute $x = 3$ into $\frac{dy}{dx}$: $\frac{dy}{dx} \bigg|_{x=3} = \frac{2}{3} - \frac{1}{2 \cdot 3^2} = \frac{2}{3} - \frac{1}{18} = \frac{12 - 1}{18} = \frac{11}{18}$ So, the slope of $t_1$ (and therefore of $t_2$, since they are parallel) is $\frac{11}{18}$.

3. Find the y-coordinate at $x = 3$: Substitute $x = 3$ into the original equation: $y = \frac{2}{3} \cdot 3 - \frac{7}{2} + \frac{1}{2 \cdot 3} = 2 - \frac{7}{2} + \frac{1}{6}$ Simplify: $y = 2 - 3.5 + 0.1667 = -1.3333$ So, the point $(3, -1.3333)$ is on the curve.

4. Equation of the tangent line $t_2$: Use the point-slope form $y - y_1 = m(x - x_1)$: $y + 1.3333 = \frac{11}{18} (x - 3)$ Simplify to get the equation of $t_2$: $y = \frac{11}{18}x - \frac{11}{6} - 1.3333 = \frac{11}{18}x - 2.8333$

Question 4

Differentiate each expression with respect to $x$.

(a) $\frac{3x^2 - 6}{2\sqrt{x}}$

1. Rewrite the expression: $f(x) = \frac{3x^2 - 6}{2x^{1/2}}$
2. Apply the quotient rule: Let $u = 3x^2 - 6$ and $v = 2x^{1/2}$. $u' = 6x, \quad v' = x^{-1/2}$ Differentiate using the quotient rule: $f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}$

(b) $3x^{\frac{3}{2}} + \frac{\sqrt{8}}{x}$

Differentiate each term separately: 1.