Final answer:
To find the inverse Laplace transform of f1(s) = 2s^2 - 8s + 16, you can use the inverse Laplace transform table. The inverse Laplace transform of f1(s) is 2t^2 - 8t + 16.
Step-by-step explanation:
To find the inverse Laplace transform of f1(s) = 2s^2 - 8s + 16, we can use the inverse Laplace transform table.
Let's break down f1(s) into its individual terms:
- L^-1(2s^2) = 2t^2
- L^-1(-8s) = -8t
- L^-1(16) = 16
Now, combine these individual inverse Laplace transforms to find the overall inverse Laplace transform of f1(s):
L^-1(f1(s)) = 2t^2 - 8t + 16