(i) Sketch the graphs of \(y = e^{-\frac{1}{2}x}\) and \(y = 4 - x^2\). The intersection points of these graphs represent the roots of the equation. The graph \(y = e^{-\frac{1}{2}x}\) is an exponential decay curve, while \(y = 4 - x^2\) is a downward-opening parabola. The graphs intersect at one positive and one negative point, indicating one positive root and one negative root.

(ii) Calculate the values at \(x = -1\) and \(x = -1.5\):

For \(x = -1\):

\(e^{-\frac{1}{2}(-1)} = e^{0.5} \approx 1.6487\)

\(4 - (-1)^2 = 3\)

Since \(1.6487 > 3\), the function changes sign between \(-1\) and \(-1.5\).

For \(x = -1.5\):

\(e^{-\frac{1}{2}(-1.5)} = e^{0.75} \approx 2.1170\)

\(4 - (-1.5)^2 = 1.75\)

Since \(2.1170 > 1.75\), the function changes sign between \(-1\) and \(-1.5\).

(iii) Use the iterative formula \(x_{n+1} = -\sqrt{4 - e^{-\frac{1}{2}x_n}}\):

Start with an initial guess, say \(x_0 = -1\).

\(x_1 = -\sqrt{4 - e^{-\frac{1}{2}(-1)}} \approx -1.4142\)

\(x_2 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4142)}} \approx -1.4106\)

\(x_3 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4106)}} \approx -1.4100\)

\(x_4 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4100)}} \approx -1.4100\)

The iterations converge to \(-1.41\) to 2 decimal places.