To analyze the function $f(x) = \frac{x}{x+1}$, we will determine if it is injective (one-to-one), surjective (onto), and find its inverse.

1. Injectivity

A function is injective if different inputs yield different outputs. We check this by assuming $f(a) = f(b)$ for $a \neq b$:

$\frac{a}{a+1} = \frac{b}{b+1}$

Cross-multiplying gives:

$a(b+1) = b(a+1) \implies ab + a = ab + b \implies a = b$

Since $a = b$ implies injectivity, $f(x)$ is injective.

2. Surjectivity

A function is surjective if every possible output is achieved by some input. The range of $f(x)$ can be analyzed as follows:

$y = \frac{x}{x+1}$

Rearranging gives:

$y(x+1) = x \implies yx + y = x \implies x - yx = y \implies x(1 - y) = y \implies x = \frac{y}{1-y}$

This shows that for every $y$ in the range $y \neq 1$, there exists an $x$. The function approaches $1$ as $x \to \infty$ and as $x \to -1$ from the left. Therefore, the range is all real numbers except $1$.

Thus, $f(x)$ is not surjective onto $\mathbb{R}$ because $1$ is not achieved.

3. Finding the Inverse

From the rearrangement earlier, we found:

$f^{-1}(y) = \frac{y}{1-y}$

Summary

• Injective: Yes
• Surjective: No (onto $\mathbb{R} \setminus \{1\}$)
• Inverse: $f^{-1}(y) = \frac{y}{1-y}$

If you have any questions or need further details, let me know!

Related Questions:

1. What are the implications of a function being injective or surjective?
2. How would the graph of $f(x)$ look?
3. What is the significance of the limit of $f(x)$ as $x \to \infty$?
4. How would you find the domain of $f^{-1}(y)$?
5. What other functions can be analyzed similarly for injectivity and surjectivity?

Tip:

To confirm injectivity, consider checking whether the derivative of the function is always positive or always negative over its domain.