Key Concepts and Formulas

• Surjective Function: A function $f: D \to A$ is surjective (or onto) if for every element $y$ in the codomain $A$, there exists at least one element $x$ in the domain $D$ such that $f(x) = y$. This means the range of the function is equal to its codomain.
• Range of a Function: The set of all possible output values of a function.
• Inequality Solving: Techniques for solving inequalities, such as the sign-chart method for rational inequalities.
• Properties of Real Numbers: For any real number $x$, $x^2 \ge 0$.

Step-by-Step Solution

1. Understand the Problem and Goal: We are given a function $f(x) = \frac{x^2}{1 - x^2}$ with domain $\mathbb{R} - \{1, -1\}$ and codomain $A$. We are told that the function is surjective, which means its codomain $A$ must be equal to its range. Our objective is to find the range of $f(x)$.

2. Set $f(x) = y$ to find the range: To determine the possible values of $y$, we set the function equal to $y$ and attempt to solve for $x$ in terms of $y$. $y = \frac{x^2}{1 - x^2}$

3. Express $x^2$ in terms of $y$: We need to isolate $x^2$ to use the property that $x^2 \ge 0$. Multiply both sides by $(1 - x^2)$: $y(1 - x^2) = x^2$ Distribute $y$: $y - yx^2 = x^2$ Move terms involving $x^2$ to one side: $y = x^2 + yx^2$ Factor out $x^2$: $y = x^2(1 + y)$ Solve for $x^2$: $x^2 = \frac{y}{1 + y}$

4. Apply the condition $x^2 \ge 0$: Since $x$ is a real number from the domain $\mathbb{R} - \{1, -1\}$, $x^2$ must be non-negative. Therefore, the expression for $x^2$ must satisfy: $\frac{y}{1 + y} \ge 0$

5. Solve the inequality $\frac{y}{1 + y} \ge 0$: We use the sign-chart method. The critical points are where the numerator or denominator is zero. Numerator: $y = 0$ Denominator: $1 + y = 0 \implies y = -1$

These critical points divide the number line into three intervals: $(-\infty, -1)$, $(-1, 0)$, and $(0, \infty)$.

• Interval $(-\infty, -1)$: Choose $y = -2$. $\frac{-2}{1 + (-2)} = \frac{-2}{-1} = 2 \ge 0$. This interval is included.
• Interval $(-1, 0)$: Choose $y = -0.5$. $\frac{-0.5}{1 + (-0.5)} = \frac{-0.5}{0.5} = -1 < 0$. This interval is excluded.
• Interval $(0, \infty)$: Choose $y = 1$. $\frac{1}{1 + 1} = \frac{1}{2} \ge 0$. This interval is included.

Now consider the critical points:

• $y = -1$: The denominator is zero, so the expression is undefined. Thus, $y = -1$ is excluded.
• $y = 0$: The numerator is zero, $\frac{0}{1+0} = 0 \ge 0$. Thus, $y = 0$ is included.

Combining these, the solution to the inequality is $y \in (-\infty, -1) \cup [0, \infty)$.

6. Check for domain restrictions on $x$: The domain of $f$ excludes $x=1$ and $x=-1$, which means $x^2 \ne 1$. Let's see if our derived expression for $x^2$ can be equal to 1. If $x^2 = 1$, then $\frac{y}{1 + y} = 1$. This leads to $y = 1 + y$, which simplifies to $0 = 1$, a contradiction. This confirms that $x^2$ can never be 1, so our range calculation automatically respects the domain restrictions on $x$.

7. Determine the codomain A: Since the function is surjective, its codomain $A$ is equal to its range. Therefore, $A = (-\infty, -1) \cup [0, \infty)$. This set can be expressed as all real numbers except for the interval $[-1, 0)$. So, $A = \mathbb{R} - [-1, 0)$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful when manipulating inequalities. Multiplying by a term that can be negative can flip the inequality sign, which is why the sign-chart method is preferred for rational inequalities.
• Endpoint Inclusion/Exclusion: Remember that when solving $P(y)/Q(y) \ge 0$, the roots of the numerator are included if the inequality is non-strict ($\ge$ or $\le$), but the roots of the denominator are always excluded.
• Domain vs. Range: Clearly distinguish between the domain of the function (given restrictions on $x$) and the range of the function (derived restrictions on $y$).

Summary

For a surjective function, the codomain is equal to the range. We found the range of $f(x) = \frac{x^2}{1 - x^2}$ by setting $f(x) = y$, solving for $x^2$ in terms of $y$, and applying the condition $x^2 \ge 0$. This led to the inequality $\frac{y}{1 + y} \ge 0$, whose solution is $y \in (-\infty, -1) \cup [0, \infty)$. This set represents the range of the function and therefore the codomain $A$.

The final answer is $\boxed{\text{R – [–1, 0)}}$.