The inverse Laplace transform of each of the following functions needs to be determined: H(s) = se⁻⁴ˢ/(3s+2)(s-2).Solution: The Laplace transform of a function f(t) is represented by F(s) which is defined as: F(s) = ∫₀^∞e^-st * f(t) dtThe inverse Laplace transform is a mathematical operation that is used to convert F(s) back into f(t).To find the inverse Laplace transform of H(s), we use partial fraction expansion. Let's write H(s) as follows:H(s) = A/(3s + 2) + B/(s - 2) + CsWhere A, B, and C are constantsMultiplying both sides by the denominator of H(s), we get: se^-4s = A(s - 2)Cs + B(3s + 2)Cs + (3s + 2)(s - 2)CSetting s = 2, we obtain: 2e^-8A = 0 => A = 0Setting s = -2/3, we get: -2/3 e^4/3 B = 0 => B = 0Now we are left with: se^-4s = Cs(3s - 2)(s - 2)After solving for C, we have: C = 1/12Thus: H(s) = 1/12(1/(3s + 2) + 3/(s - 2))The inverse Laplace transform of H(s) is therefore:H(t) = 1/12e^-2(t-4)u(t-4) + 1/4e^2(t-4)u(t-4) (where u(t) is the unit step function)Answer: The inverse Laplace transform of H(s) = se⁻⁴ˢ/(3s+2)(s-2) is H(t) = 1/12e^-2(t-4)u(t-4) + 1/4e^2(t-4)u(t-4).