(b) Determine an equation of the line tangent to the graph of \( f \) at \( (a, f(a)) \) for the given value of \( a \). i) \( f(x)=\sqrt{3 x+1} ; \quad a=8 \) ii) \( f(x)=\frac{2}{3 x+1} ; \quad a=-1 \)

Sure, let's determine the equations of the tangent lines for both functions at the specified points. --- ### **i) \( f(x) = \sqrt{3x + 1} \) at \( a = 8 \)** **Step 1: Find \( f(8) \)** \[ f(8) = \sqrt{3(8) + 1} = \sqrt{24 + 1} = \sqrt{25} = 5 \] So, the point of tangency is \( (8, 5) \). **Step 2: Find the derivative \( f'(x) \)** \[ f(x) = (3x + 1)^{1/2} \\ f'(x) = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}} \] **Step 3: Evaluate \( f'(8) \)** \[ f'(8) = \frac{3}{2\sqrt{3(8) + 1}} = \frac{3}{2\sqrt{25}} = \frac{3}{10} \] So, the slope \( m = \frac{3}{10} \). **Step 4: Write the equation of the tangent line using the point-slope form** \[ y - f(a) = m(x - a) \\ y - 5 = \frac{3}{10}(x - 8) \] **Simplified Equation:** \[ y = \frac{3}{10}x + \frac{13}{5} \] --- ### **ii) \( f(x) = \frac{2}{3x + 1} \) at \( a = -1 \)** **Step 1: Find \( f(-1) \)** \[ f(-1) = \frac{2}{3(-1) + 1} = \frac{2}{-3 + 1} = \frac{2}{-2} = -1 \] So, the point of tangency is \( (-1, -1) \). **Step 2: Find the derivative \( f'(x) \)** \[ f(x) = 2(3x + 1)^{-1} \\ f'(x) = 2 \cdot (-1)(3x + 1)^{-2} \cdot 3 = \frac{-6}{(3x + 1)^2} \] **Step 3: Evaluate \( f'(-1) \)** \[ f'(-1) = \frac{-6}{3(-1) + 1)^2} = \frac{-6}{(-3 + 1)^2} = \frac{-6}{4} = -\frac{3}{2} \] So, the slope \( m = -\frac{3}{2} \). **Step 4: Write the equation of the tangent line using the point-slope form** \[ y - f(a) = m(x - a) \\ y - (-1) = -\frac{3}{2}(x - (-1)) \\ y + 1 = -\frac{3}{2}(x + 1) \] **Simplified Equation:** \[ y = -\frac{3}{2}x - \frac{5}{2} \] --- ### **Final Answers:** 1. **For \( f(x) = \sqrt{3x + 1} \) at \( a = 8 \):** \[ y = \frac{3}{10}x + \frac{13}{5} \] 2. **For \( f(x) = \frac{2}{3x + 1} \) at \( a = -1 \):** \[ y = -\frac{3}{2}x - \frac{5}{2} \]