Understanding the average slope of a curve

Question

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?

Answer 1

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.