a. Find the derivative function \( f^{\prime} \) for the function \( f \). b. Determine an equation of the line tangent to the graph of \( f \) at \( (a, f(a)) \) for the given value of a. \( f(x)=\sqrt{3 x+4}, a=4 \) a. \( f^{\prime}(x)=\square \)

To find the derivative function \( f' \) for the function \( f(x) = \sqrt{3x + 4} \), we will use the chain rule. ### a. Finding the derivative \( f' \) 1. Rewrite the function in exponent form: \[ f(x) = (3x + 4)^{1/2} \] 2. Apply the chain rule: \[ f'(x) = \frac{1}{2}(3x + 4)^{-1/2} \cdot (3) \] Here, we differentiate the outer function and multiply by the derivative of the inner function \( 3x + 4 \). 3. Simplify: \[ f'(x) = \frac{3}{2\sqrt{3x + 4}} \] Thus, the derivative function is: \[ f'(x) = \frac{3}{2\sqrt{3x + 4}} \] ### b. Finding the equation of the tangent line at \( (a, f(a)) \) where \( a = 4 \) 1. First, calculate \( f(4) \): \[ f(4) = \sqrt{3(4) + 4} = \sqrt{12 + 4} = \sqrt{16} = 4 \] So the point is \( (4, 4) \). 2. Next, calculate \( f'(4) \): \[ f'(4) = \frac{3}{2\sqrt{3(4) + 4}} = \frac{3}{2\sqrt{16}} = \frac{3}{2 \cdot 4} = \frac{3}{8} \] 3. Now, use the point-slope form of the equation of a line: \[ y - f(a) = f'(a)(x - a) \] Substituting \( a = 4 \), \( f(4) = 4 \), and \( f'(4) = \frac{3}{8} \): \[ y - 4 = \frac{3}{8}(x - 4) \] 4. Rearranging to get the equation in slope-intercept form: \[ y - 4 = \frac{3}{8}x - \frac{3}{2} \] \[ y = \frac{3}{8}x - \frac{3}{2} + 4 \] \[ y = \frac{3}{8}x - \frac{3}{2} + \frac{8}{2} \] \[ y = \frac{3}{8}x + \frac{5}{2} \] Thus, the equation of the tangent line at the point \( (4, 4) \) is: \[ y = \frac{3}{8}x + \frac{5}{2} \]