To find the range of the function \( f(x) = \frac{x \lfloor x \rfloor}{x^2 + 1} \) defined on the interval \( (1, 3) \), we will analyze the function in two parts based on the greatest integer function \( \lfloor x \rfloor \).
### Step 1: Determine the intervals for \( \lfloor x \rfloor \)
1. For \( x \in (1, 2) \):
- \( \lfloor x \rfloor = 1 \)
- Thus, \( f(x) = \frac{x \cdot 1}{x^2 + 1} = \frac{x}{x^2 + 1} \)
2. For \( x \in (2, 3) \):
- \( \lfloor x \rfloor = 2 \)
- Thus, \( f(x) = \frac{x \cdot 2}{x^2 + 1} = \frac{2x}{x^2 + 1} \)
### Step 2: Analyze \( f(x) \) for \( x \in (1, 2) \)
For \( x \in (1, 2) \):
\[
f(x) = \frac{x}{x^2 + 1}
\]
To find the maximum and minimum values, we differentiate \( f(x) \):
\[
f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}
\]
Setting \( f'(x) = 0 \):
\[
1 - x^2 = 0 \implies x = 1 \text{ or } x = -1
\]
Since \( x \in (1, 2) \), we only consider \( x = 1 \).
- Evaluate \( f(x) \) at the endpoints:
- \( f(1) = \frac{1}{1^2 + 1} = \frac{1}{2} \)
- \( f(2) = \frac{2}{2^2 + 1} = \frac{2}{5} \)
Since \( f'(x) \) changes sign from positive to negative, \( f(x) \) is decreasing in \( (1, 2) \). Thus, the range in this interval is:
\[
\left( \frac{2}{5}, \frac{1}{2} \right)
\]
### Step 3: Analyze \( f(x) \) for \( x \in (2, 3) \)
For \( x \in (2, 3) \):
\[
f(x) = \frac{2x}{x^2 + 1}
\]
Differentiating \( f(x) \):
\[
f'(x) = \frac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}
\]
Setting \( f'(x) = 0 \):
\[
2 - 2x^2 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1
\]
Again, only \( x = 2 \) is relevant in this interval.
- Evaluate \( f(x) \) at the endpoints:
- \( f(2) = \frac{4}{5} \)
- \( f(3) = \frac{6}{10} = \frac{3}{5} \)
Since \( f'(x) \) changes sign from positive to negative, \( f(x) \) is decreasing in \( (2, 3) \). Thus, the range in this interval is:
\[
\left( \frac{3}{5}, \frac{4}{5} \right)
\]
### Step 4: Combine the ranges
Combining the ranges from both intervals, we have:
\[
\text{Range of } f(x) = \left( \frac{2}{5}, \frac{1}{2} \right) \cup \left( \frac{3}{5}, \frac{4}{5} \right)
\]
### Final Answer:
The range of \( f(x) \) is:
\[
\left( \frac{2}{5}, \frac{1}{2} \right) \cup \left( \frac{3}{5}, \frac{4}{5} \right)
\]
