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Let's analyze the given problem step by step.

### Functions:
- \( f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \), defined as \( f(x) = \frac{1}{x} \)
- \( g: \mathbb{R} \to \mathbb{R} \), defined as \( g(x) = \frac{2x}{1 + x^2} \)

### Part (a):
We are asked to determine which of the two functions, \( f \) or \( g \), is **injective** (one-to-one) and **surjective** (onto). We also need to find the image of \( g(\mathbb{R}) \), i.e., the range of \( g(x) \).

#### For \( f(x) = \frac{1}{x} \):
1. **Injectivity (One-to-one)**: 
   - The function \( f(x) = \frac{1}{x} \) is injective because if \( f(x_1) = f(x_2) \), then \( \frac{1}{x_1} = \frac{1}{x_2} \), which implies \( x_1 = x_2 \).
   
2. **Surjectivity (Onto)**:
   - The function \( f(x) = \frac{1}{x} \) is surjective, because for any real number \( y \), there exists an \( x = \frac{1}{y} \) such that \( f(x) = y \). The domain excludes 0, so every real number is covered.

Thus, \( f(x) \) is both injective and surjective.

#### For \( g(x) = \frac{2x}{1 + x^2} \):
1. **Injectivity (One-to-one)**: 
   - To check injectivity, consider \( g(x_1) = g(x_2) \), which gives \( \frac{2x_1}{1 + x_1^2} = \frac{2x_2}{1 + x_2^2} \). Solving this equation leads to \( x_1 = x_2 \), so \( g(x) \) is injective.
   
2. **Surjectivity (Onto)**:
   - The range of \( g(x) = \frac{2x}{1 + x^2} \) is bounded between -1 and 1. This can be checked by finding the maximum and minimum values of the function. The function achieves these values at \( x = 1 \) and \( x = -1 \), respectively. So, \( g(\mathbb{R}) = (-1, 1) \).

Thus, \( g(x) \) is injective but not surjective.

### Part (b):
We need to determine the composition \( h = g \circ f \), which means we are finding \( h(x) = g(f(x)) = g\left(\frac{1}{x}\right) \).

Substituting \( f(x) \) into \( g(x) \):
\[
h(x) = g\left(\frac{1}{x}\right) = \frac{2 \left( \frac{1}{x} \right)}{1 + \left( \frac{1}{x} \right)^2} = \frac{\frac{2}{x}}{1 + \frac{1}{x^2}} = \frac{\frac{2}{x}}{\frac{x^2 + 1}{x^2}} = \frac{2x}{x^2 + 1}
\]

Thus, the composition \( h(x) \) is the same as \( g(x) \), meaning \( h(x) = \frac{2x}{x^2 + 1} \).

### Summary:
- (a) The function \( f(x) = \frac{1}{x} \) is both injective and surjective. The function \( g(x) = \frac{2x}{1 + x^2} \) is injective but not surjective, with \( g(\mathbb{R}) = (-1, 1) \).
- (b) The composition \( h(x) = g(f(x)) = \frac{2x}{x^2 + 1} \).

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Do you want further details or have any questions about the steps? Here are five related questions that can help expand your understanding:

1. How can we use the derivative to verify the injectivity of a function?
2. What method could be used to find the maximum and minimum values of \( g(x) \)?
3. How would the result change if we defined \( f(x) \) for \( x = 0 \)?
4. Can a function be injective but not surjective? Provide an example.
5. How do we determine the composition of two functions in general?

**Tip**: To check injectivity, one useful approach is the horizontal line test for graphical functions or solving \( f(x_1) = f(x_2) \) algebraically.

Given two functions f(x) = 1/x (defined for all real numbers except 0) and g(x) = 2x/(1 + x^2), determine which function is injective and surjective. Also, find g(ℝ) and determine the composition of h = g(f(x)).

Analyze the injectivity and surjectivity of functions f(x) = 1/x and g(x) = 2x/(1 + x^2). Explore the function composition h(x) = g(f(x)) and determine the range of g(x).

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