A time-domain signal $$f(t)$$ is represented by the following expression in the frequency domain:

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

The final value of $$f(t)$$ is most nearly:

In order to solve this problem, we must utilize the Final Value Theorem (F.V.T) which is listed on the Laplace Transform Pair table on pg. 56 of our FE Reference Handbook below:

Therefore:

$$\begin{gathered} \underset{t\to \infty }{lim}f\left(t\right) = \underset{s\to 0}{lim}sF\left(s\right) \\ \underset{t\to \infty }{lim}f\left(t\right) = \underset{s\to 0}{lim}s\left(\frac{1}{s^2+9s}\right) \\ \underset{t\to \infty }{lim}f\left(t\right) = \underset{s\to 0}{lim}s\left(\frac{1}{s(s+9)}\right) \\ \underset{t\to \infty }{lim}f\left(t\right) = \underset{s\to 0}{lim}\frac{1}{s+9} = \frac{1}{0+9}=\frac{1}{9} \end{gathered}$$

According to the theorem, our final value of f(t) is therefore approximately $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{9}$}\right.$.