If a function `f(x)` increases in the interval `(a, b).` then the function `phi(x) = [f(x)]^n` increases in the same interval and `phi(x) != f(x)` if

Options

A. `n=-1`
B. `n=0`
C. `n=3 `
D. `n=4 `

Answer

To solve the problem, we need to analyze the function \( \phi(x) = [f(x)]^n \) under the condition that \( f(x) \) is an increasing function in the interval \( (a, b) \). We want to determine for which values of \( n \) the function \( \phi(x) \) is also increasing, and we need to ensure that \( \phi(x) \neq f(x) \).

### Step 1: Understand the condition of increasing functions
Since \( f(x) \) is increasing in the interval \( (a, b) \), we know that:
\[ f'(x) > 0 \quad \text{for all } x \in (a, b) \]

### Step 2: Differentiate \( \phi(x) \)
The function \( \phi(x) \) is defined as:
\[ \phi(x) = [f(x)]^n \]
To determine if \( \phi(x) \) is increasing, we need to find its derivative:
\[ \phi'(x) = n [f(x)]^{n-1} f'(x) \]

### Step 3: Analyze the sign of \( \phi'(x) \)
For \( \phi(x) \) to be increasing, we need:
\[ \phi'(x) > 0 \]
This leads us to the inequality:
\[ n [f(x)]^{n-1} f'(x) > 0 \]

### Step 4: Consider the components of the inequality
- Since \( f'(x) > 0 \) (because \( f(x) \) is increasing), the sign of \( \phi'(x) \) depends on \( n \) and \( [f(x)]^{n-1} \).
- If \( f(x) > 0 \) in the interval, then \( [f(x)]^{n-1} > 0 \) for \( n-1 \) being even or \( n \) being odd.

### Step 5: Determine conditions for \( n \)
1. **If \( n > 0 \)**: Both \( n \) and \( [f(x)]^{n-1} \) are positive, thus \( \phi'(x) > 0 \).
2. **If \( n = 0 \)**: \( \phi(x) = 1 \) (a constant function), which is neither increasing nor decreasing.
3. **If \( n < 0 \)**: \( [f(x)]^{n-1} \) becomes problematic. For example, if \( n = -1 \), \( \phi(x) = \frac{1}{f(x)} \), which is decreasing since \( f(x) \) is increasing.

### Step 6: Evaluate the options
- **Option 1**: \( n = -1 \) → \( \phi(x) \) is decreasing.
- **Option 2**: \( n = 0 \) → \( \phi(x) \) is constant.
- **Option 3**: \( n = 3 \) → \( \phi(x) \) is increasing.
- **Option 4**: \( n = 4 \) → \( \phi(x) \) is increasing, but we cannot conclude definitively without knowing the sign of \( f(x) \).

### Conclusion
The function \( \phi(x) = [f(x)]^n \) is guaranteed to be increasing for \( n = 3 \) and potentially for \( n = 4 \) depending on the sign of \( f(x) \). However, since we need \( \phi(x) \neq f(x) \), the best option is \( n = 3 \).

### Final Answer
Thus, the value of \( n \) for which \( \phi(x) \) is increasing and \( \phi(x) \neq f(x) \) is:
\[ \boxed{3} \]