Answer
False
Work Step by Step
When $t \to 4^{-}$, then $f(t)=(4)^2=16$
When $t \to 4^{+}$, then $f(t)=(4)^2=16$ and When $t =4$, then $f(t)=4$
We see that $f(t)_{-}=f(t)_{+} \ne f(t)$
Therefore, the given statement is $\bf{False}$.
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.1 Definition of the Laplace Transform - True-False Review - Page 675: a
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