How was Hubble Space Telescope deployed?

We are given, $v(t)=0.001302t^3-0.09029t^2+23.61t-3.083$ $\therefore\text{ acceleration } a(t)=\frac{\delta{v}}{\delta{t}}=0.003906x^2-0.18058x+23.61 \quad\frac{\text{feet}}{s^2}$

Observe:
$\rightarrow$ $a(t)$ is a proper polynomial function (continuous, differentiable), but its domain is a closed interval .
$\rightarrow$ So, its (local and global) extreme's can be only at points where its derivative is $0$ and at the boundary values (domain is a closed interval, remember!) .
$\rightarrow$ So, if we evaluate the function at all such points and compare the result with each other, we'd have found the global maximum and minimum values.

$a'(t)=0.007821x-0.18058$ $a'(t)=0\implies x=\frac{0.18058}{0.007821}=23.089119038486125$ $a(23.089119038486125)=21.522884458634206$ $a(0)=23.61\quad\ldots\text{(the lower bound value of domain)}$ $a(126) = 62.868576000000004 \quad \ldots \text{ ( the upper bound value of domain )}$ $\therefore l_1=62 \text{ \& }l_2=21\implies l_1+l_2=\boxed{83}$

The acceleration is a quadratic function, $a(t) = 0.003906\:t^2 - 0.18058\:t + 23.61.$ Since the quadratic coefficient is positive, this function has precisely one minimum, at the vertex, where $t_\star = \frac{-(-0.18058)}{2\cdot 0.003906} = 23.12\ \text{s}.$ This lies in the given interval ( $0 \leq 23.12 \leq 126$ ), so this is where the minimum acceleration happens. Evaluate $a_\star = a(t_\star) = 21.523\ \text{ft/s}^2\ \ \therefore\ \ I_2 = 21.$ The maximum must occur at the edges of the domain. Clearly, $a(0) = 23.61$ , but this is less than $a(126) = 62.87\ \text{ft/s}^2\ \ \therefore\ \ I_1 = 62.$ The desired answer is therefore $I_1 + I_2 = \boxed{83}$ .