If L { f (t)} = F(s) = s3+1(s−1)2(s2+2s+5) , find the following:(i) f (t) (ii) L {t f (t)} (iii) L ©e−2t f (t)ª (iv) L−1 ©e−3sL { f (t)}

The problem you've given is related to Laplace Transforms, a mathematical tool often used in engineering and physics. Here's how you can solve it:

(i) To find f(t), you need to find the inverse Laplace transform of F(s). However, F(s) is not in a standard form that can be easily recognized from a table of Laplace transforms. You need to perform partial fraction decomposition on F(s) to express it as a sum of simpler fractions. Once you've done that, you can look up the inverse Laplace transforms of each of these simpler fractions in a table of Laplace transforms to find f(t).

(ii) To find L{t f(t)}, you need to use the property of Laplace transforms that L{t f(t)} = -dF(s)/ds, where F(s) is the Laplace transform of f(t). So, you need to find the derivative of F(s) with respect to s, and then change the sign.

(iii) To find L{e^(-2t) f(t)}, you need to use the property of Laplace transforms that L{e^(at) f(t)} = F(s-a), where F(s) is the Laplace transform of f(t). So, you need to replace every s in F(s) with s+2.

(iv) To find L^(-1){e^(-3s) L{f(t)}}, you need to use the property of Laplace transforms that L^(-1){e^(-as) F(s)} = f(t-a) u(t-a), where F(s) is the Laplace transform of f(t), L^(-1) is the inverse Laplace transform, and u(t-a) is the unit step function. So, you need to find f(t-3) u(t-3), where f(t) is the function you found in part (i).

Please note that without the actual partial fraction decomposition of F(s), I can't provide the specific functions for f(t), L{t f(t)}, L{e^(-2t) f(t)}, and L^(-1){e^(-3s) L{f(t)}}.