(i) Evaluate the function at \(x = 1\) and \(x = 2\):

\(f(1) = 1^5 - 3(1)^3 + 1^2 - 4 = 1 - 3 + 1 - 4 = -5\)

\(f(2) = 2^5 - 3(2)^3 + 2^2 - 4 = 32 - 24 + 4 - 4 = 8\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(ii) Rearrange the equation:

Start with \(x^5 - 3x^3 + x^2 - 4 = 0\).

Add 4 to both sides: \(x^5 - 3x^3 + x^2 = 4\).

Divide by \(x^2\): \(x^3 - 3x + \frac{1}{x} = \frac{4}{x^2}\).

Rearrange to: \(x^3 = 3x + \frac{4}{x^2} - 1\).

Take the cube root: \(x = \sqrt[3]{3x + \frac{4}{x^2} - 1}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{3x_n + \frac{4}{x_n^2} - 1}\).

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt[3]{3(1.5) + \frac{4}{(1.5)^2} - 1} = 1.8041\)

\(x_2 = \sqrt[3]{3(1.8041) + \frac{4}{(1.8041)^2} - 1} = 1.7817\)

\(x_3 = \sqrt[3]{3(1.7817) + \frac{4}{(1.7817)^2} - 1} = 1.7800\)

\(x_4 = \sqrt[3]{3(1.7800) + \frac{4}{(1.7800)^2} - 1} = 1.7800\)

The root correct to 2 decimal places is 1.78.