To find the range of the function \( f(x) = \sqrt{x^2 - 5x} \), we can start by analyzing the expression inside the square root, \( x^2 - 5x \), since the square root function requires that the expression inside it must be non-negative (i.e., \( x^2 - 5x \geq 0 \)).

### Step 1: Solve the inequality \( x^2 - 5x \geq 0 \)

1. Factor the quadratic expression:
   \[
   x^2 - 5x = x(x - 5)
   \]
2. Set the inequality:
   \[
   x(x - 5) \geq 0
   \]

This is a product of two factors. To solve this inequality, we analyze the sign of the product by considering the critical points where \( x(x - 5) = 0 \). These critical points are \( x = 0 \) and \( x = 5 \).

3. Determine the sign in each interval:

   - For \( x < 0 \), both factors \( x \) and \( x - 5 \) are negative, so their product \( x(x - 5) \) is positive.
   - For \( 0 < x < 5 \), \( x \) is positive and \( x - 5 \) is negative, so their product \( x(x - 5) \) is negative.
   - For \( x > 5 \), both factors \( x \) and \( x - 5 \) are positive, so their product \( x(x - 5) \) is positive.

Therefore, the inequality \( x(x - 5) \geq 0 \) holds when \( x \leq 0 \) or \( x \geq 5 \).

### Step 2: Determine the range of \( f(x) \)

Given that \( f(x) = \sqrt{x^2 - 5x} \), the function is defined only for \( x \leq 0 \) or \( x \geq 5 \). Let's evaluate the function at the boundary points:

- At \( x = 0 \):
  \[
  f(0) = \sqrt{0^2 - 5(0)} = \sqrt{0} = 0
  \]
- At \( x = 5 \):
  \[
  f(5) = \sqrt{5^2 - 5(5)} = \sqrt{25 - 25} = \sqrt{0} = 0
  \]

Now, analyze the behavior of the function as \( x \) tends towards positive and negative infinity:

- For \( x \geq 5 \), as \( x \) increases, \( x^2 - 5x \) grows larger, making \( f(x) \) also increase without bound.
- For \( x \leq 0 \), as \( x \) becomes more negative, \( x^2 - 5x \) grows larger as well, so \( f(x) \) increases without bound in the negative direction.

### Conclusion

The function \( f(x) = \sqrt{x^2 - 5x} \) is zero at \( x = 0 \) and \( x = 5 \), and it increases without bound as \( x \) moves away from these points in either direction.

Thus, the range of \( f(x) \) is:
\[
\text{Range of } f(x) = [0, \infty)
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How would you find the domain of the function \( g(x) = \sqrt{4x - x^2} \)?
2. What is the range of \( f(x) = \sqrt{9 - x^2} \)?
3. Can you determine the vertex form of a quadratic function and its significance?
4. How do you solve inequalities involving square roots?
5. What happens to the range if the function is modified to \( f(x) = \sqrt{x^2 - 5x + 6} \)?

**Tip:** Always check the domain of the expression inside a square root before determining the range, as it directly influences where the function is defined.

find the range f(x) = square root x^2 - 5x

Explore how to determine the range of the function f(x) = √(x^2 - 5x) with a detailed, step-by-step solution. This problem delves into algebraic concepts, including quadratic functions and inequalities, making it suitable for students in Grades 9-12.

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