Question

Explain why the function

$\mathbf{f}\left(t\right)\mathbf{=}\{\begin{array}{ll}t,& 0\le t<2\\ 4,& 2<t<5\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(t-5\right)$}\right.,& t>5\end{array}$

is not piecewise continuous on$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}\mathbf{.}$

Short answer

The given function is not piecewise continuous since it is not continuous on the last interval. This is due to the fact that there is a vertical asymptote in the last interval.

Step-by-step solution

Definition of Laplace Transform

Let $\mathbf{f}$be a function defined for $\mathbf{t}\mathbf{\ge }\mathbf{0}$. Then the integral

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

is said to be the Laplace transform of $\mathbf{f}$, provided that the integral converges.

Use Laplace Transform

The given function is not piecewise continuous since it is not continuous on the last interval. This is due to the fact that there is a vertical asymptote in the last interval. Also, we require that the piecewise function is defined at one of the subinterval endpoints.