Key Concepts and Formulas

• Range of a Function: The set of all possible output values of a function.
• Odd Function: A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. The graph of an odd function is symmetric with respect to the origin.
• AM-GM Inequality: For non-negative real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$, with equality if and only if $a=b$.
• Algebraic Manipulation for Range: Setting $y = f(x)$ and solving for $x$ in terms of $y$. The values of $y$ for which $x$ is real constitute the range.
• Limits: Understanding the behavior of a function as $x$ approaches specific values (including infinity).

Step-by-Step Solution

Step 1: Analyze Function Properties

We are given the function $f(x) = \frac{x}{1+x^2}$.

• Domain: The denominator $1+x^2$ is always positive for all real $x$ ($x^2 \ge 0 \implies 1+x^2 \ge 1$). Thus, the domain of $f(x)$ is $R$.
• Symmetry: We check if $f(x)$ is an odd or even function: $f(-x) = \frac{-x}{1+(-x)^2} = \frac{-x}{1+x^2} = - \left( \frac{x}{1+x^2} \right) = -f(x)$ Since $f(-x) = -f(x)$, $f(x)$ is an odd function. This means its graph is symmetric about the origin. If we find the range for $x \ge 0$, we can infer the range for $x < 0$.
• Value at $x=0$: $f(0) = \frac{0}{1+0^2} = 0$ So, 0 is in the range of $f(x)$.

Step 2: Determine the Range for $x > 0$ using AM-GM Inequality

For $x > 0$, we can rewrite $f(x)$ by dividing the numerator and denominator by $x$: $f(x) = \frac{x/x}{(1+x^2)/x} = \frac{1}{\frac{1}{x} + \frac{x^2}{x}} = \frac{1}{\frac{1}{x} + x}$ Now, we apply the AM-GM inequality to the terms in the denominator, $x$ and $\frac{1}{x}$, which are both positive for $x > 0$: $\frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}}$ $\frac{x + \frac{1}{x}}{2} \ge \sqrt{1}$ $\frac{x + \frac{1}{x}}{2} \ge 1$ $x + \frac{1}{x} \ge 2$ This inequality tells us that the minimum value of the denominator $x + \frac{1}{x}$ is 2, and this occurs when $x = \frac{1}{x}$, which implies $x^2 = 1$. Since we are considering $x > 0$, this happens at $x=1$.

Now, we can find the upper bound for $f(x)$ for $x > 0$: Since $x + \frac{1}{x} \ge 2$, taking the reciprocal of both sides (and reversing the inequality sign) gives: $\frac{1}{x + \frac{1}{x}} \le \frac{1}{2}$ Thus, for $x > 0$, $f(x) \le \frac{1}{2}$. The maximum value of $\frac{1}{2}$ is achieved at $x=1$, as $f(1) = \frac{1}{1+1^2} = \frac{1}{2}$.

To find the lower bound for $f(x)$ when $x > 0$, we consider the limit as $x$ approaches 0 from the positive side: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{x + \frac{1}{x}}$ As $x \to 0^+$, $x \to 0$ and $\frac{1}{x} \to \infty$. So, $x + \frac{1}{x} \to \infty$. $\lim_{x \to 0^+} f(x) = \frac{1}{\infty} = 0$ Also, consider the limit as $x \to \infty$: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{x + \frac{1}{x}}$ As $x \to \infty$, $x \to \infty$ and $\frac{1}{x} \to 0$. So, $x + \frac{1}{x} \to \infty$. $\lim_{x \to \infty} f(x) = \frac{1}{\infty} = 0$ Since $f(x) = \frac{1}{x + \frac{1}{x}}$ and $x + \frac{1}{x} > 0$ for $x > 0$, $f(x)$ will always be positive. The function approaches 0 but never reaches it for $x > 0$. Therefore, for $x > 0$, the range of $f(x)$ is $(0, \frac{1}{2}]$.

Step 3: Determine the Range for $x < 0$ using Symmetry

Since $f(x)$ is an odd function, for every value $y$ in the range for $x > 0$, the value $-y$ will be in the range for $x < 0$. The range for $x > 0$ is $(0, \frac{1}{2}]$. So, for $x < 0$, the range will be $[-\frac{1}{2}, 0)$.

Step 4: Combine the Ranges and State the Final Range

We have found:

• For $x > 0$, the range is $(0, \frac{1}{2}]$.
• For $x < 0$, the range is $[-\frac{1}{2}, 0)$.
• At $x=0$, $f(0) = 0$.

Combining these sets, the total range of $f(x)$ for all real $x$ is the union of these intervals: $\left[ -\frac{1}{2}, 0 \right) \cup \{0\} \cup \left( 0, \frac{1}{2} \right] = \left[ -\frac{1}{2}, \frac{1}{2} \right]$

Alternative Method: Algebraic Manipulation (Setting $y=f(x)$)

Let $y = f(x) = \frac{x}{1+x^2}$. To find the range, we solve for $x$ in terms of $y$: $y(1+x^2) = x$ $y + yx^2 = x$ $yx^2 - x + y = 0$ This is a quadratic equation in $x$. For $x$ to be a real number, the discriminant ($\Delta$) must be non-negative.

Case 1: $y = 0$. If $y=0$, the equation becomes $0 \cdot x^2 - x + 0 = 0$, which simplifies to $-x = 0$, so $x=0$. Since $x=0$ is a real number, $y=0$ is in the range.

Case 2: $y \ne 0$. For the quadratic equation $yx^2 - x + y = 0$ to have real solutions for $x$, the discriminant must be greater than or equal to zero: $\Delta = (-1)^2 - 4(y)(y) \ge 0$ $1 - 4y^2 \ge 0$ $1 \ge 4y^2$ $\frac{1}{4} \ge y^2$ $y^2 \le \frac{1}{4}$ Taking the square root of both sides: $-\sqrt{\frac{1}{4}} \le y \le \sqrt{\frac{1}{4}}$ $-\frac{1}{2} \le y \le \frac{1}{2}$ Combining this result with Case 1 (where $y=0$ is included), the range of $f(x)$ is $[-1/2, 1/2]$.

Common Mistakes & Tips

• Forgetting the $y=0$ case: When using the discriminant method for $yx^2 - x + y = 0$, it's crucial to check if the coefficient of $x^2$ (which is $y$) can be zero. If $y=0$, the equation is linear, not quadratic, and requires separate analysis.
• Mistakes with AM-GM equality: Remember that the equality in AM-GM holds only when the terms are equal. This is essential for finding the exact maximum or minimum values of the function.
• Not considering the domain: Always ensure that the values of $x$ obtained from solving for $y$ are within the domain of the original function. In this case, the domain is all real numbers, so this is not an issue.

Summary

We determined the range of the function $f(x) = \frac{x}{1+x^2}$ by first analyzing its properties, specifically its odd symmetry. This allowed us to focus on finding the range for positive $x$ values. Using the AM-GM inequality on the transformed expression $f(x) = \frac{1}{x + \frac{1}{x}}$, we established that for $x > 0$, $0 < f(x) \le \frac{1}{2}$. Due to the odd symmetry, for $x < 0$, the range is $-\frac{1}{2} \le f(x) < 0$. Including the value $f(0)=0$, the complete range of the function is $[-1/2, 1/2]$. The algebraic method of setting $y=f(x)$ and solving for $x$ also confirmed this result by analyzing the discriminant of the resulting quadratic equation.

The final answer is $\boxed{\left[ { - {1 \over 2},{1 \over 2}} \right]}$.