Animated Solution for Mathematics - Functions: If the function $f:\mathbb{R}-\{1,-1\}\to A$ defined by $f(x)=\frac{x^2}{1-x^2}$, is surjective, then $A$ is equal to

We are examining the function $f(x)=\frac{x^2}{1-x^2}$. Our objective is to determine the codomain $A$ that renders this function surjective.

A function is surjective if and only if its codomain is equal to its range. Therefore, our task is to identify the set of all possible values $y$ that the function can produce.

We begin by setting $y=\frac{x^2}{1-x^2}$. To understand the constraints on $y$, we isolate $x$ through algebraic manipulation.

Cross-multiplying gives us:

Expanding and grouping the $x^2$ terms:

Solving for $x^2$, we obtain the master equation:

For $x$ to be a real number, we must satisfy the condition $x^2\ge 0$. Substituting our expression for $x^2$, we arrive at the following inequality:

This inequality defines the boundary of our output space. Any value of $y$ that violates this condition would require $x$ to be imaginary, which is not permitted in the real domain.

To solve $\frac{y}{1+y}\ge 0$, we identify the critical points where the expression changes sign. These occur at $y=0$ (from the numerator) and $y=-1$ (from the denominator).

Applying the wavy curve method: 1. For $y>0$, the expression is positive. 2. For $-1<y<0$, the expression is negative. 3. For $y<-1$, the expression is positive.

Including the point $y=0$ (where the expression is zero) and excluding $y=-1$ (where the expression is undefined), the range is $(-\infty ,-1)\cup [0,\infty )$.

The range of the function is $(-\infty ,-1)\cup [0,\infty )$. To express this in the form $\mathbb{R}-S$, we identify the values excluded from the real number line.

The values missing from the range are those in the interval $