Answer
$$
F(t)=-10.28+175.9 t e^{-t / 1.3}
$$
(a)
first use the product rule to find the derivative, $F^{\prime}(t)$
$$
\begin{aligned}
F^{\prime}(t)=&(175.9)\left(e^{-t / 13}\right) \\
&+(175.99 t)\left(-\frac{1}{1.3} e^{-t / 13}\right) \\
=&(175.9)\left(e^{-t / 1.3}\right)\left(1-\frac{t}{1.3}\right) \\
\approx & 175.9 e^{-t / 1.3}(1-0.769 t)
\end{aligned}
$$
(b)
The function $F(t) $ is increasing on $ (0, 1.3 ).$
The function $F(t) $ is decreasing on $(1.3 , \infty).$
Work Step by Step
$$
F(t)=-10.28+175.9 t e^{-t / 1.3}
$$
(a)
first use the product rule to find the derivative, $F^{\prime}(t)$
$$
\begin{aligned}
F^{\prime}(t)=&(175.9)\left(e^{-t / 13}\right) \\
&+(175.99 t)\left(-\frac{1}{1.3} e^{-t / 13}\right) \\
=&(175.9)\left(e^{-t / 1.3}\right)\left(1-\frac{t}{1.3}\right) \\
\approx & 175.9 e^{-t / 1.3}(1-0.769 t)
\end{aligned}
$$
To find any intervals where this function is increasing or decreasing, set $F^{\prime}(t)=0$
$$
\begin{aligned}
F^{\prime}(t) &=175.9 e^{-t / 1.3}(1-0.769 t)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\\
& 1-0.769 t=0\\
& -0.769 t=-1\\
\Rightarrow\quad\quad\quad\quad\quad\\\
& t=\frac{-1}{-0.769}\\
& t\approx 1.30039 \\
\end{aligned}
$$
so $1.3$ is the only critical number.
Now, we can use the first derivative test, since the domain is $(0,\infty )$. Check the sign of $F^{\prime}(t)$ in the intervals:
$$
(0, 1.3 ), \quad (1.3, \infty),
$$
(1)
Test a number in the interval $(0, 1.3 ) $ say $1$:
$$
\begin{aligned}
F^{\prime}(1) &=175.9 e^{-(1) / 1.3}(1-0.769 (1)) \\
&=18.82804 \\
& \gt 0
\end{aligned}
$$
we see that $F^{\prime}(t)$ is positive in that interval, so $F(t)$ is increasing on $(0, 1.3)$.
(2)
Test a number in the interval $(1.3 , \infty ) $ say $2$:
$$
\begin{aligned}
F^{\prime}(2) &=175.9 e^{-(2) / 1.3}(1-0.769 (2)) \\
&=-20.31902 \\
&\lt 0
\end{aligned}
$$
we see that $F^{\prime}(t)$ is negative in that interval, so $F(t)$ is decreasing on $(1.3 , \infty )$.
So,
(b)
The function $F(t) $ is increasing on $ (0, 1.3 ).$
The function $F(t) $ is decreasing on $(1.3 , \infty).$
