Calculate this average rate of change to show that it is the same as the slope m. f(x2) − f(x1) x2 − x1

To show that the average rate of change of the function $f(x) = mx + b$ is equal to the slope $m$, we can calculate the average rate of change between two points $x_1$ and $x_2$.

The average rate of change is given by:

$\frac{f(x_2) - f(x_1)}{x_2 - x_1}$

Substitute $f(x) = mx + b$ into the formula:

$\frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{(mx_2 + b) - (mx_1 + b)}{x_2 - x_1}$

Simplify the numerator:

$(mx_2 + b) - (mx_1 + b) = mx_2 + b - mx_1 - b$

Combine like terms:

$mx_2 + b - mx_1 - b = mx_2 - mx_1$

Factor out $m$:

$mx_2 - mx_1 = m(x_2 - x_1)$

Substitute back into the formula for the average rate of change:

$\frac{m(x_2 - x_1)}{x_2 - x_1}$

Simplify by canceling $x_2 - x_1$ (assuming $x_2 \neq x_1$):

$m$

Thus, the average rate of change of the function $f(x) = mx + b$ over any interval $[x_1, x_2]$ is equal to the slope $m$.

If you have any further questions or need more details, feel free to ask!

Here are some related questions to explore further:

1. How does the average rate of change differ for a quadratic function compared to a linear function?
2. Can the average rate of change be zero? What does that signify?
3. What happens to the average rate of change when $x_1 = x_2$?
4. How can we interpret the average rate of change geometrically?
5. How would you find the instantaneous rate of change for a function?

Tip: The average rate of change of a function over an interval is geometrically the slope of the secant line connecting the two endpoints of the interval on the function's graph.