Question

suppose f(t) is a function for which f’(t) is piecewise continuous and exponential order c. use result in this section and section

Justify$\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{s}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$

Where $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathit{L}\left\{f\left(t\right)\right\}$verify this result with$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{k}\mathit{t}\mathbf{.}$

Short answer

f(t) is a function f’(t) is piecewise continuous and justify exponential order c.

$\begin{array}{rcl}L\left\{f\text{'}\left(t\right)\right\}& =& sF\left(s\right)-f\left(0\right)\\ F\left(s\right)& =& L\left\{f\left(t\right)\right\}\\ \underset{x\to \infty }{\lim }sF\left(s\right)& =& f\left(0\right)\end{array}$

Step-by-step solution

Step1: definition of piecewise continuous.

piecewise function is continuous on a given interval in its domain: its constituent functions are continuous on the corresponding intervals (subdomains), and there is no discontinuity at each endpoint of the subdomains within that interval.

{Theorem 7.2.2 }$ If f and f’ are continuous on and are of exponential order, and if f’(t) is piecewise continuous on , then

$\begin{array}{rcl}L\left\{f\text{'}\left(t\right)\right\}& =& sF\left(s\right)-f\left(0\right)\\ F\left(s\right)& =& L\left\{f\left(t\right)\right\}\end{array}$

{theorem } $ if piecewise continuous on$\left[0,\infty \right)$

There exponential order, $F\left(s\right)=L\left\{f\left(t\right)\right\}$,then

$\underset{s\to \infty }{\lim }F\left(s\right)=0$

Now, the function f’ satisfies the conditions of the function f in theorem $\ textbf {theorem $7.1.3$}$, so we have

data-custom-editor="chemistry" $$\begin{gathered} \underset{x\to \infty }{\lim }L\left\{f\text{'}\left(t\right)\right\}=0 \\ \Downarrow \left(1\right) \\ \underset{x\to \infty }{\lim }\left[sF\left(s\right)-f\left(0\right)\right]=0 \end{gathered}$$

$\underset{s\to \infty }{\lim }\left[sF\left(s\right)\right]=0$

Step2: u.se result in this section

$\begin{array}{rcl}\underset{s\to \infty }{\lim }\left[s\frac{s}{s^2+k^2}\right]& =& \cos \theta \\ \underset{s\to \infty }{\lim }\left[\frac{s^2}{\left(1+\frac{k^2}{s^2}\right)}\right]& =& 1\\ \underset{s\to \infty }{\lim }\left[\frac{1}{1+\frac{k^2}{s^2}}\right]& =& 1\\ \frac{1}{1+0}& =& 1\\ 1& =& 1\end{array}$

Hence,

$\underset{s\to \infty }{\lim }\left[sF\left(s\right)\right]=f\left(0\right)$