(a) Sketch the graphs of \(y = \ln x\) and \(y = 3x - x^2\). The graph of \(y = \ln x\) passes through (1, 0) and has the y-axis as an asymptote. The graph of \(y = 3x - x^2\) is a parabola opening downwards, symmetric about the line \(x = 1.5\), passing through (0, 0) and (3, 0). The intersection point indicates one real root.

(b) Calculate \(\ln 2 \approx 0.693\) and \(3(2) - 2^2 = 2\). Since \(0.693 < 2\), the function changes sign between 2 and 2.8. Calculate \(\ln 2.8 \approx 1.029\) and \(3(2.8) - 2.8^2 = 1.04\). Since \(1.029 < 1.04\), the root lies between 2 and 2.8.

(c) Use the iterative formula \(x_{n+1} = \sqrt{3x_n - \ln x_n}\) starting with \(x_1 = 2.5\):

\(x_2 = \sqrt{3(2.5) - \ln(2.5)} \approx 2.6178\)

\(x_3 = \sqrt{3(2.6178) - \ln(2.6178)} \approx 2.6311\)

\(x_4 = \sqrt{3(2.6311) - \ln(2.6311)} \approx 2.6300\)

The root is approximately 2.63 to 2 decimal places.