linear approximation to the velocity suppose I have a table with information of a person who is jumping on a trampoline. and I am given nothing more than a table.
$$\begin{array}{c|c|c|} 
t (sec) & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} \\ \hline
\text{x(t) (feet)} & 2.5 & 3.6 & 1.2 & 2.7 & 4.4 \\ \hline
\end{array}$$ 
I need to find the linear approximation to the velocity of the jumper at $t_1=3.75$. the equation for linear approximation is $L(t_1)=x(t)+x'(t)(t_1-t)$. How do I have to find the $x(t)$? It is going to be an instant velocity, but I cannot take the limit, since i need the exact number.
and then I am asked to find the best estimation for the acceleration at $t=3.7$
Any help appreciated.
 A: Edit: Looking over the question again, the entire exercise seems nonsensical. We are given $x$ coordinates in feet, which implies that $x$ is a distance measurement. Linear interpolation of distance measurements gives you segments of constant velocity (zero acceleration) with instants of undefined acceleration between segments.
To ask for the acceleration at $t = 3.7$ is either trivial or requires something beyond linear interpolation of the given data.
On top of that, the given data cannot possibly be the data for an actual person jumping on a trampoline unless they are just snapshots of the height at those instants, interleaved with bounces in between, and the bounces make linear interpolation invalid.
If we change "feet" to "feet/second" then $x$ is a velocity. If positive $x$ represents upward velocity, however, then to increase the velocity (as we do between $t=3$ an $t=4$ there has to be a bounce in between the times, which again invalidates linear interpolation between those two times.
If we ignore the framing story about the trampoline and just blindly interpolate the data linearly, we can estimate the height $x$ at any time on the assumption of a piecewise constant velocity. But this seems unsatisfactory. Unless the original problem is severely misrepresented here,
it seems fatally flawed.
The following is my original answer:

Based on the formula, it appears you are supposed to do piecewise linear
interpolation, which means that you are approximating the function $x(t)$
with a function whose graph consists of multiple line segments, one
segment connecting each consecutive pair of points where the
value of $x(t)$ is known.
So to apply the linear approximation formula $L(t_1)=x(t)+x'(t)(t_1-t)$
given a set of values of $t$ and $x(t)$, there are really only two
values of $t$ that will be of interest to you: the greatest value less than $t_1$
and the least value greater than $t_1$.
In other words, try to insert $t_1 = 3.75$ in the proper numeric sequence in
the first row of your table; you can see it fits between $3$ and $4$,
so those are the only two columns of the table that matter
when interpolating at $3.75$.
So you have
\begin{align}
x(3) & = 1.2 \\
x(4) & = 2.7
\end{align}
and the assumption of linear interpolation is that between those two points,
your estimated $x(t)$ will be the same as a
function whose graph is a straight line through those two points.
So if the graph of $x(t)$ were a straight line through those two points,
what would its slope be?
Try
$$
x' = \frac{2.7 - 1.2}{4 - 3}.
$$
Since the slope of a straight line is constant, that's our estimate
of $x'$ for any $t$.
So let $t = 3$, for example, and you have
$$L(t_1) = x(t)+x'(t)(t_1-t) = x(3) + x'(3)(t_1 - 3)
= 1.2 + \frac{2.7 - 1.2}{4 - 3}(t_1 - 3).$$
You could also try $t = 4$. If you work this out completely, it will come
to the same answer.
If you are later given another value $t_2$ to interpolate that is not between
$3$ and $4$, then you have to throw away all the numbers we
just calculated and start over with some other pair of values of $t$,
because now you're on a part of the graph which is a line segment between
a different pair of points, and it (usually) lies on a different straight line.
A: According to your formula for Linear Approximation:
$$L(t_1)=x(t)-x'(t)(t_1-t)$$
Now, we set $t_1  = 3.75$. For $t$, I would suggest setting $t=4$, since $3.75$ is closest to $4$ on the table.
Therefore, for your question, we get that:
$$L(3.75) = x(4) - x'(4)(3.75-4) = 2.7 -\frac{2.7}{4}*(-0.25) = 2.86875$$
Now to estimate the acceleration, use the velocities at $t=3.75$ and $t=4$:
$$A(3.7) =  \frac{V(4) - V(3.75)}{4-3.75} = \frac{0.675 - 2.86875}{0.25} = -8.775$$
