F s( ) is the Laplace transform of the function

F(1) is ________ (correct to two decimal places).

The correct answer is 0.5.

Based on the provided image, the function is given by:
$f\left(t\right)=2t^2{e}^{-t}$

We want to find $F\left(s\right)$, which is the Laplace transform of $f\left(t\right)$, and evaluate it at $s=1$.

First, recall the standard Laplace transform formula for $t^n$:
$L\left\{t^n\right\}=\frac{n!}{s^{n+1}}$

For $n=2$, this gives:
$L\left\{t^2\right\}=\frac{2!}{s^3}=\frac{2}{s^3}$

Next, we apply the frequency shifting property of the Laplace transform, which states that:
$L\left\{e^{at}g\right(t\left)\right\}=G(s-a)$

In our case, $a=-1$, so the transform of $t^2{e}^{-t}$ becomes:
$L\left\{t^2{e}^{-t}\right\}=\frac{2}{{(s+1)}^3}$

Therefore, the Laplace transform $F\left(s\right)$ of $f\left(t\right)=2t^2{e}^{-t}$ is:
$F\left(s\right)=2·\frac{2}{{(s+1)}^3}=\frac{4}{{(s+1)}^3}$

Now, we evaluate $F\left(s\right)$ at $s=1$:
$F\left(1\right)=\frac{4}{{(1+1)}^3}=\frac{4}{2^3}=\frac{4}{8}=0.5$