Final answer:
The steady-state gain of the given Transfer Function, C(t) = 5 - ½ e^{-2t} + 8 e^{-t}, is the value as t approaches infinity, which results in the exponential terms decaying to zero, leaving the constant term which is 5. None of the options provided (-7.5, 7.5, 7, -7) match this value.
Step-by-step explanation:
The question asks for the steady-state gain of a Transfer Function (T/F) in time constant form when given a function C(t) = 5 - ½ e^{-2t} + 8 e^{-t}. The steady-state gain is the value of C(t) as t approaches infinity. In this case, the exponential terms e^{-2t} and e^{-t} decay to zero as t increases because the base e is the base of the natural logarithm and it represents the decay constant for the nuclide, and the term ½ e^{-2t} and 8 e^{-t} represent two components of the function that decrease exponentially over time. This results in C(t) approaching the constant term, which is 5.
Therefore, the steady-state gain of the T/F function is 5. Considering the options given, and since there is no option '5', it appears there may be a typo in the options or in the understanding of the question. As for the steady-state gain of the function, it is 5, not any of the options given, which are -7.5, 7.5, 7, or -7.