I have been asked to compute the Laplace Transform of the following piecewise function
\begin{equation} f(t) = \begin{cases} t - 1 \quad 1 \leq t < 2 \\ 3 - t \quad 2 \leq t < 3 \\ 0 \quad t \geq 3 \end{cases} \end{equation}
I could not find an example where the domain of the piecewise function to transform does not begin at $0$, I was wondering if it could be possible to shift the function to the left $f(t + 1)$, so that the domain begins at $0$, rewrite in terms of the unit step function and transform it. Is that something valid to do? or how should these cases be handled?
Set $b(t)=u(t)-u(t-1)$ as the unit box or square function with support $[0,1]$. Then your function can be represented as $$ f(t)=(b*b)(t-1), $$ check it that the convolution of $b$ with itself gives the hat function. Now apply standard Laplace rules $$ \begin{align} {\cal L}\{f(t)\}(s)&=e^{-s}{\cal L}\{(b*b)(t)\}\\ &=e^{-s}\Bigl({\cal L}\{b(t)\}\Bigr)^2 \end{align} $$ etc.