find the average rate of change of y=$$ \sqrt{2x-1} $$ $$  $$between x=1 and x=5 

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.

Learn how to calculate the average rate of change of the function y = √(2x - 1) between x = 1 and x = 5 using the Average Rate of Change formula. Understand the concepts of square root functions and intervals in calculus. Suitable for advanced high school students.

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