If `f:RtoR,f(x)=(sqrt(x^(2)+1)-3x)/(sqrt(x^(2)+1)+x)` then find the range of `f(x)`.

Explanation

To find the range of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by

\[ f(x) = \frac{\sqrt{x^2 + 1} - 3x}{\sqrt{x^2 + 1} + x}, \]

we will follow these steps:

### Step 1: Analyze the function
The function is defined for all real numbers \( x \) since the denominator \( \sqrt{x^2 + 1} + x \) is never zero (as \( \sqrt{x^2 + 1} \) is always positive and \( x \) can be negative, but their sum is always positive).

### Step 2: Find the derivative
To determine the behavior of the function, we will find the derivative \( f'(x) \) using the quotient rule:

\[ f'(x) = \frac{(u'v - uv')}{v^2} \]

where \( u = \sqrt{x^2 + 1} - 3x \) and \( v = \sqrt{x^2 + 1} + x \).

Calculating \( u' \) and \( v' \):

- \( u' = \frac{x}{\sqrt{x^2 + 1}} - 3 \)
- \( v' = \frac{x}{\sqrt{x^2 + 1}} + 1 \)

Now applying the quotient rule:

\[ f'(x) = \frac{(\frac{x}{\sqrt{x^2 + 1}} - 3)(\sqrt{x^2 + 1} + x) - (\sqrt{x^2 + 1} - 3x)(\frac{x}{\sqrt{x^2 + 1}} + 1)}{(\sqrt{x^2 + 1} + x)^2} \]

### Step 3: Simplify the derivative
After simplifying the expression for \( f'(x) \), we find that \( f'(x) < 0 \) for all \( x \). This indicates that \( f(x) \) is a decreasing function.

### Step 4: Evaluate the limits at infinity
Next, we evaluate the limits of \( f(x) \) as \( x \) approaches \( -\infty \) and \( +\infty \):

1. **As \( x \to -\infty \)**:

\[ f(x) = \frac{\sqrt{x^2 + 1} - 3x}{\sqrt{x^2 + 1} + x} \approx \frac{-3x}{-3x} = 1. \]

2. **As \( x \to +\infty \)**:

\[ f(x) = \frac{\sqrt{x^2 + 1} - 3x}{\sqrt{x^2 + 1} + x} \approx \frac{-2x}{2x} = -1. \]

### Step 5: Determine the range
Since \( f(x) \) is a decreasing function, and it approaches \( 1 \) as \( x \to -\infty \) and \( -1 \) as \( x \to +\infty \), the range of \( f(x) \) is:

\[ (-1, 1). \]

### Final Answer
Thus, the range of the function \( f(x) \) is:

\[ \text{Range of } f(x) = (-1, 1). \]