Using the table Laplace transform, determine the function f (t) of the following functions F (s).

F(s)= 1/s^2 + 3s/s^2+9

F(s)= 6/s^2 + 10s+29

F(s)= 1/2s + 8/s^2+4s+8 - 3/s^2

To determine the function f(t) from the given F(s), we need to apply the inverse Laplace transform. The inverse Laplace transform converts a function from the s-domain to the t-domain.

1. F(s) = 1/s^2 + 3s/s^2+9
To find f(t), we need to perform the inverse Laplace transform on F(s). Using the table of Laplace transforms, we can break down F(s) into simpler terms whose inverse transforms are known.

1/s^2 transform: The inverse transform of 1/s^2 is t.

3s/(s^2+9) transform: The inverse transform of 3s/(s^2+9) is 3cos(3t).

Therefore, f(t) = t + 3cos(3t).

2. F(s) = 6/s^2 + 10s + 29
Applying the inverse Laplace transform to F(s), we break it down into known transforms.

6/s^2 transform: The inverse transform of 6/s^2 is 6t.

10s transform: The inverse transform of 10s is 10.

29 transform: The inverse transform of 29 is 29.

Therefore, f(t) = 6t + 10 + 29 = 6t + 39.

3. F(s) = 1/2s + 8/(s^2+4s+8) - 3/s^2
Applying the inverse Laplace transform to F(s), we break it down into known transforms.

1/2s transform: The inverse transform of 1/2s is (1/2)ln(t).

8/(s^2+4s+8) transform:
The denominator s^2+4s+8 can be factored into (s+2)^2+4.
The inverse transform of 8/(s^2+4s+8) is 4e^(-2t)sin(2t).

-3/s^2 transform: The inverse transform of -3/s^2 is -3t.

Therefore, f(t) = (1/2)ln(t) + 4e^(-2t)sin(2t) - 3t.

In summary:
- For F(s) = 1/s^2 + 3s/s^2+9, f(t) = t + 3cos(3t).
- For F(s) = 6/s^2 + 10s + 29, f(t) = 6t + 39.
- For F(s) = 1/2s + 8/s^2+4s+8 - 3/s^2, f(t) = (1/2)ln(t) + 4e^(-2t)sin(2t) - 3t.