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.