To find the average rate of change of the function $y = \sqrt{2x - 1}$ over the interval $[x_1, x_2]$, you can use the formula:

$\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

Given the interval $[1, 5]$, we have:

• $x_1 = 1$
• $x_2 = 5$

The function is $y = \sqrt{2x - 1}$, so we need to find $f(1)$ and $f(5)$:

$f(1) = \sqrt{2(1) - 1} = \sqrt{2 - 1} = \sqrt{1} = 1$ $f(5) = \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3$

Now, plug these values into the formula:

$\text{Average Rate of Change} = \frac{f(5) - f(1)}{5 - 1} = \frac{3 - 1}{5 - 1} = \frac{2}{4} = \frac{1}{2}$

Thus, the average rate of change of $y = \sqrt{2x - 1}$ between $x = 1$ and $x = 5$ is $\frac{1}{2}$.

Would you like further details or have any questions about this calculation?

Here are 5 numbered questions for you to ask:

1. How do you find the average rate of change for a quadratic function?
2. What is the geometric interpretation of the average rate of change?
3. How do you find the derivative of $\sqrt{2x - 1}$?
4. Can you explain the concept of a secant line in calculus?
5. How do you solve equations involving square roots?

Tip: When working with square root functions, remember to consider the domain of the function to ensure the expressions under the square root are non-negative. This helps avoid dealing with complex numbers when focusing on real-valued functions.