To find the average rate of change of the function $f(x) = e^{x - 1} - 3$ between $x = 1$ and $x = 3$, we use the formula for the average rate of change over an interval $[a, b]$:

$\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}$

Here, $a = 1$, $b = 3$, and $f(x) = e^{x - 1} - 3$.

1. Evaluate $f(1)$: $f(1) = e^{1 - 1} - 3 = e^0 - 3 = 1 - 3 = -2$

2. Evaluate $f(3)$: $f(3) = e^{3 - 1} - 3 = e^2 - 3$

3. Compute the average rate of change: $\text{Average Rate of Change} = \frac{f(3) - f(1)}{3 - 1} = \frac{(e^2 - 3) - (-2)}{2}$

   Simplifying further: $= \frac{e^2 - 3 + 2}{2} = \frac{e^2 - 1}{2}$

Thus, the average rate of change of $f(x) = e^{x - 1} - 3$ from $x = 1$ to $x = 3$ is:

$\frac{e^2 - 1}{2}$

Would you like further details, or do you have any questions?

Here are some related questions to consider:

1. How does the average rate of change differ for other intervals, such as $x = 2$ to $x = 4$?
2. What is the derivative of $f(x) = e^{x - 1} - 3$, and how does it relate to the instantaneous rate of change?
3. Can the average rate of change be negative for other intervals of this function?
4. How does changing the constant term $-3$ in $f(x)$ affect the average rate of change?
5. How would this calculation differ if $f(x)$ were a polynomial instead of an exponential function?

Tip: The average rate of change for exponential functions often varies significantly across intervals due to the rapid growth rate of exponential terms.