The function $f : (-\infty, \infty) \to (-\infty, 1)$, defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$, is:

Step 1: Simplify the function

First, we rewrite $2^{-x}$ as $\frac{1}{2^x}$ to simplify the expression. This makes it easier to combine the terms in the numerator and the denominator.

Step 2: Combine terms

Next, we find a common denominator for the terms in the numerator and the denominator. After combining them, we can cancel out the common factor of $2^x$ from both the numerator and the denominator, resulting in a simpler form of the function.

Step 3: Analyze the range of the function

To understand the range of the function, we can rewrite it by adding and subtracting 1 in the numerator. This transformation allows us to express the function as $1 - \frac{2}{2^{2x} + 1}$, which makes it easier to analyze its behavior.

Step 4: Determine the behavior of $2^{2x}$

The term $2^{2x}$ is always positive for any real value of $x$. As $x \to \infty$, $2^{2x} \to \infty$. As $x \to -\infty$, $2^{2x} \to 0$.

Step 5: Determine the range of $f(x)$

When $x$ approaches infinity, $2^{2x} + 1$ approaches infinity, causing the fraction $\frac{2}{2^{2x} + 1}$ to approach 0. Thus, $f(x)$ approaches $1 - 0 = 1$. When $x$ approaches negative infinity, $2^{2x} + 1$ approaches 1, causing the fraction $\frac{2}{2^{2x} + 1}$ to approach 2. Thus, $f(x)$ approaches $1 - 2 = -1$. Since $2^{2x} + 1$ is always greater than 1, the fraction $\frac{2}{2^{2x} + 1}$ is always positive and less than 2. Therefore, $f(x)$ is always between -1 and 1.

Step 6: Check for injectivity (one-to-one)

To check if the function is injective (one-to-one), we assume $f(x_1) = f(x_2)$ and show that this implies $x_1 = x_2$. After cross-multiplication and simplification, we find that $2^{2x_1} = 2^{2x_2}$, which means $2x_1 = 2x_2$, and therefore $x_1 = x_2$. This confirms that the function is injective.

Step 7: Check for surjectivity (onto)

To check for surjectivity (onto), we need to see if for every $y$ in the codomain $(-\infty, 1)$, there exists an $x$ in the domain $(-\infty, \infty)$ such that $f(x) = y$. We solve for $x$ in terms of $y$. The expression for $x$ is defined only when $\frac{y+1}{1-y} > 0$. This inequality holds when $-1 < y < 1$. Since the given codomain is $(-\infty, 1)$, and our range is $(-1, 1)$, the function is not surjective onto $(-\infty, 1)$. However, if the codomain was $(-1, 1)$, it would be surjective.

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