Unfortunately, $$p = \frac{ f'\left(0\right) + f'\left(x\right)}{2}$$ does not give the average rate of change. For example, try $f(x)=1-\cos x$. Your formula gives the average rate of change from $0$ to $\pi$ as $0$, when instead it should clearly be positive from examining the graph.
This is like saying that the average value of a function $f(x)$ on an interval $[a,b]$ is $\frac{f(a)+f(b)}{2}$. This is untrue for most functions. It happens to be true for linear functions, which are exactly what you get when taking the derivative of a quadratic polynomial, explaining why it works in that case.
You are certainly not alone in wondering about this! I should ask: in what sense do you mean the question?
a) If your question -- "can a specific moment in time really have a rate of change?" -- is directed towards the physical world, and the words "time" or "moment" are to be taken as referring to those things from our daily experience, then I'd tell you not to forget what math does: it doesn't constitute the real world, it just models it.
Perhaps the space we live in is actually discrete; i.e. if you zoom in close enough, our world is made of atomic "cells", just like a Minecraft world. Suppose each cell is a cube $1.6 \times 10^{-45}$ meters (ten orders of magnitude below the Planck length) on an edge. We don't know if this hypothesis is true or not: what experiment would disprove it? If it were true, then some things about real numbers that we learn in math (i.e. the idea of the limit is based, that for any number you name, I can always name a smaller one*), would be "wrong" for talking about objects on that size scale.
But it would still work just as well, as an approximation, for things that we currently use calculus for -- e.g. to calculate where to aim our spaceships. The rocket equations themselves are never going to fit the situation exactly (have you accounted for that dust particle? and that one?), the numbers we put into them are never going to be measured precisely.
A model cannot be judged right or wrong in itself; only the application of a model to a real-world situation can be judged, and then only in grades -- more appropriate or less appropriate. If speed comes in discrete chunks, then there may be no moment at which the volleyball, whose arc is described by $y = -x^2$, is ever moving at $-4$ meters/second calculus would predict at $x = 2$. Or maybe speed is continuous, and there is such a moment.
There's no way, even in principle, to tell, so we stick with the model we've got and change it only when it predicts the real world incorrectly.
b) But being less high-minded, it's helpful to have several ways to think about these things (and don't let anyone, including me, convince you that you have to think only their way about it).
As others have said, the derivative of a function $f(x)$ is a function $f'(x)$ which gives you the slope of the tangent line at $x$. If you believe that there can be a tangent line at a single point, then you can just think of that when others say "instantaneous rate of change".
*Here's the technical definition of a limit (ripped from Wikipedia), in case it helps. The statement
$$
\lim_{x \rightarrow 0} f(x) = L
$$
means that you can make $f(x)$ as close to $L$ as you like by making $x$ sufficiently close to $0$. With variables, that's:
For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x| < \delta$, then $|f(x) - L| < \epsilon$.
You can see how this would not work if there was a smallest real number -- then if I choose $\epsilon$ equal to that number, how are you going to make $|f(x) - L|$ smaller than it?
Best Answer
The average rate of change is the slope of the line connecting the two endpoints at $(-3, f(-3))$ and $(0, f(0))$.
It is givn by $$\frac{f(0)-f(-3)}{0-(-3)}$$
The instantaneous rate of change is the derivative $4e^{\cos x}$.
Now you must find the $x$ such that the instanteous rate of change of $h(x)$ equals the average change of $f(x)$ over the interval. Hopefully there is such an $x$!