Which of the following functions is/are bounded?

Options

A. `f(x)=(2x)/(1+x^(2)),[-2,2]`
B. `f(x)=(x^(2))/(1-x),x in [0,2]-[1]`
C. `f(x)=(x^(3)-8x+6)/(4x+1),[0,5]`
D. none of these

Explanation

To determine which of the given functions are bounded, we will analyze each function step by step.

### Step 1: Analyze the first function
**Function:** \( f(x) = \frac{2x}{1 + x^2} \) for \( x \in [0, 2] \)

1. **Find the range of \( f(x) \)**:
- The function is continuous on the closed interval [0, 2].
- To find the maximum and minimum values, we can check the endpoints and critical points.
- Calculate \( f(0) = \frac{2(0)}{1 + 0^2} = 0 \).
- Calculate \( f(2) = \frac{2(2)}{1 + 2^2} = \frac{4}{5} \).
- Find the derivative \( f'(x) \) and set it to zero to find critical points:
\[ f'(x) = \frac{(1 + x^2)(2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} \]
Setting \( f'(x) = 0 \) gives \( 2 - 2x^2 = 0 \) or \( x^2 = 1 \) which gives \( x = 1 \).
- Calculate \( f(1) = \frac{2(1)}{1 + 1^2} = 1 \).
- The range of \( f(x) \) on [0, 2] is from 0 to 1, hence it is bounded.

### Step 2: Analyze the second function
**Function:** \( f(x) = \frac{x^3}{1 - x} \) for \( x \in [0, 2) \)

1. **Check for boundedness**:
- As \( x \) approaches 1, \( f(x) \) tends to infinity because the denominator approaches zero.
- Therefore, \( f(x) \) is not bounded on the interval [0, 2).

### Step 3: Analyze the third function
**Function:** \( f(x) = \frac{x^3 - 8x + 6}{4x + 1} \) for \( x \in [0, 5] \)

1. **Find the range of \( f(x) \)**:
- The function is continuous on the closed interval [0, 5].
- Calculate \( f(0) = \frac{0^3 - 8(0) + 6}{4(0) + 1} = 6 \).
- Calculate \( f(5) = \frac{5^3 - 8(5) + 6}{4(5) + 1} = \frac{125 - 40 + 6}{20 + 1} = \frac{91}{21} \approx 4.33 \).
- Find critical points by calculating the derivative and setting it to zero:
\[ f'(x) = \frac{(4x + 1)(3x^2 - 8) - (x^3 - 8x + 6)(4)}{(4x + 1)^2} \]
- Analyze the critical points and endpoints to find the maximum and minimum values.
- Since the function is continuous and does not have any vertical asymptotes in the interval, it is bounded.

### Conclusion
- The first function \( f(x) = \frac{2x}{1 + x^2} \) is bounded.
- The second function \( f(x) = \frac{x^3}{1 - x} \) is not bounded.
- The third function \( f(x) = \frac{x^3 - 8x + 6}{4x + 1} \) is bounded.

### Final Answer
The bounded functions are the first and third functions.