show if f(x)=((x)/(x+1)) is injective, surjective, and what it’s inverse

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.

Explore the function f(x) = x/(x+1) to determine its injectivity, surjectivity, and to find its inverse. This comprehensive analysis confirms that f(x) is injective, not surjective onto R, and provides its inverse as f^{-1}(y) = y/(1-y). Suitable for advanced math students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation