Question

show that t square is of exponential order 3​

Answer 1

Step-by-step explanation:

For the first question, take something like

f(t)=sin(et2).

Then f is bounded (so in particular of exponential growth), but

f′(t)=2tet2cos(et2)

is not of exponential growth. (Look at the values for z=2πk, k∈Z.)

For the second question, see Jyrki's answer.

Answer 2

Answer:

$\mathrm{t}^{2}$ is of exponential order.

Step-by-step explanation:

Given: Function

To find: show that t square is of exponential order 3​

Solution:

Given $f(t)=t^{2}$ By the definition of exponential order,

$\underset{t \rightarrow \infty}{L t} e^{-s t} f(t)=\underset{t \rightarrow \infty}{L t} e^{-s t} t^{2} \\ =\underset{t \rightarrow \infty}{L t} \frac{t^{2}}{s t}\left[\frac{\infty}{\infty}\ i.e., Indeterminant form] \\\\\[Apply L'Hospital rule].$

$=\underset{t \rightarrow \infty}{L t} \frac{2 t}{s e^{s t}}\left[\frac{\infty}{\infty}\right. i.e., Indeterminant form]$

$=\underset{t \rightarrow \infty}{L t} \frac{2 t}{s e^{S t}}\left[\frac{\infty}{\infty}\right. i.e., Indeterminant form][Apply L'Hospital rule]$

$\begin{aligned}{Lt}_{t \rightarrow \infty} \frac{2}{s^{2} e^{s t}} \\&=\frac{2}{\infty}\end{aligned}$

Hence $\mathrm{t}^{2}$ is of exponential order.