This question mainly asks: Is my understanding of the average slope correct? This question is somewhat related to my previous question. However, its different from the previous question to a certain extent.
If $P(x_0,y_0)$ and $Q(x_0+\Delta x, y_0+\Delta y)$ are two points on the curve of the graph of square function ($y=x^2$), then the average slope of the curve between these two points can be given by the equation: $\bar m = 2x_0+\Delta x$. If $\Delta x$ is infinitely small, it can be neglected and we're left with $m=2x_0$ which is the formula to find the slope of a point at the curve.
Let $P(2,4)$ and $Q(2.00005, 4.000200003)$ be two points on the curve. The average slope between these two points of the curve is $4.00005$. Now, let's take 3 points which are between the curve $PQ$, and find their slopes by the formula $2x_0$. Let those 3 points be $(2.00002, 4.00008),(2.00003, 4.0000120001)$ and $(2.00004, 4.000160002)$.
Calculating the slopes at the given points,
1) $m=4$ at $(2, 4)$
2) $m=4.00004$ at $(2.00002, 4.00008)$
3) $m=4.00006$ at $(2.00003, 4.0000120001)$
4) $m=4.00008$ at $(2.00004, 4.000160002)$
5) $m=4.0001$ at $(2.00005, 4.000200003)$
Now we shall calculate the average of above values:
$m_{av} = \frac{4+4.00004+4.00006+4.00008+4.0001}{5}$
$m_{av} = 4.000056$
Conclusion:
The average slope $\bar m = 2x_0 + \Delta x$ is called so, because it's value is approximately equal to the average slope found by dividing the sum of certain points at an interval by the total no. of points, or $\bar m ≐ m_{av}$.
I know that $\Delta x=0.00005$ is not infinitely small nor negligible, but for the sake of simplicity I've used it here.
$4.00005≈4.000056$. They indeed are approximately equal. My question is: Is the way I understand the average slope correct? The way I've given above by calculating those long values?
Best Answer
I'm not clear as to what your question is, but you're conclusion is correct. You will learn shortly why $y = x^2 \implies \;\text{slope}\; \approx 2x_0 + \Delta x$, and why as $\Delta x$ grows smaller, slope $\approx 2x$ at the point $(x, x^2),$ when you learn about derivatives.
As you calculated: $$ m_{av} = 4.000056,$$ and we can compute $$\overline m = 2 x_0 + \Delta x = 2\cdot 2 + (2.00005 - 2) = 4 + .00005 = 4.00005$$
So can you see how $\overline m \approx m_{av}$?
Added: Yes, your "experiment" using this function, and calculating slopes using $m = 2x$ at points various points for which the change in $x$ is very small and dividing by the number of slopes you calculated, indicates what average slope over a small interval seeks to estimate.