Let's consider a function, \(f(x) = x^3 - 3x^2 - 4x + 12\). Prove that there is a root in the interval (-2, 0).

Step 1 ：For proving a root lies between the interval, we can use the Intermediate Value Theorem (IVT). According to the IVT, if a function \(f\) is continuous on a closed interval \([a, b]\), and \(k\) is any number between \(f(a)\) and \(f(b)\), then there is at least one number \(c\) in the interval \((a, b)\) such that \(f(c) = k\).

Step 2 ：In this case, we need to evaluate the function at the endpoints of the interval i.e., -2 and 0.

Step 3 ：At \(x = -2\), \(f(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12 = -8 - 12 + 8 + 12 = 0\).

Step 4 ：At \(x = 0\), \(f(0) = (0)^3 - 3(0)^2 - 4(0) + 12 = 12\).

Step 5 ：So, the function changes sign over the interval (-2, 0), which indicates that there is a root in the interval according to the IVT.