To determine the intervals where the function $f(x) = \sqrt{2x - 10}$ is increasing or decreasing, we follow these steps:

1. Domain of the Function

First, let's find the domain of the function because the square root function is only defined when the expression inside is non-negative.

$2x - 10 \geq 0$ $2x \geq 10$ $x \geq 5$

Thus, the domain of the function is $[5, \infty)$.

2. First Derivative

Now, to determine the intervals of increasing or decreasing behavior, we take the first derivative of the function. Using the chain rule, we have:

$f(x) = \sqrt{2x - 10} = (2x - 10)^{\frac{1}{2}}$

The derivative is: $f'(x) = \frac{1}{2}(2x - 10)^{-\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2x - 10}}$

3. Analyzing the Derivative

• $f'(x) > 0$ when $x > 5$, meaning the function is increasing in this region.
• At $x = 5$, $f'(x)$ is undefined because the square root function has a vertical tangent at this point.

Therefore, the function is:

• Increasing on the interval $(5, \infty)$
• The function is not decreasing on any interval, as $f'(x) \geq 0$ for all $x \geq 5$.

Final Answer:

• Increasing on the interval: $(5, \infty)$
• Decreasing on the interval: None

Would you like more details or further clarification?

Here are 5 related questions you might explore:

1. How do we determine the critical points of a function using its derivative?
2. What does it mean for a function to have a vertical tangent line?
3. How does the chain rule apply when differentiating composite functions?
4. What is the role of the derivative in determining concavity?
5. How can we determine if a function is concave up or concave down?

Tip: Always check the domain of the function before analyzing its derivative behavior, especially for functions with square roots or rational expressions.