Question: fe-(3+ik) g(k) = F-'[g(k)] = f(x) a. 3 + ik F-'g(k)] = f(x) b. g ( k ) = 2+i(k- 1) = g (ki) =
![\fe-(3+ik) g(k) = F-'[g(k)] = f(x) a. 3 + ik F-'g(k)]](https://s3.amazonaws.com/si.experts.images/answers/2024/06/667b297529398_573667b29750b71d.jpg)

![(ki) = ik F-' [g(k)] = f(x) C. (1 + ik) (2](https://s3.amazonaws.com/si.experts.images/answers/2024/06/667b2975eff57_573667b2975d3b85.jpg)

\fe-(3+ik) g(k) = F-'[g(k)] = f(x) a. 3 + ik F-'g(k)] = f(x) b. g ( k ) = 2+i(k- 1) = g (ki) = ik F-' [g(k)] = f(x) C. (1 + ik) (2 + ik)
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