Suppose the function `f(x)=x^(n),n!=0` is differentiable for all x. Then n can be any element of the interval

Options

A. `(1,oo)`
B. `(0,oo)`
C. `((1)/(2),oo)`
D. None of the above

Explanation

To solve the problem, we need to determine the values of \( n \) for which the function \( f(x) = x^n \) is differentiable for all \( x \).

### Step-by-Step Solution:

1. **Understanding Differentiability**:
A function is differentiable at a point if its derivative exists at that point. For the function \( f(x) = x^n \), we need to find its derivative.

2. **Finding the Derivative**:
The derivative of \( f(x) = x^n \) is given by:
\[ f'(x) = n \cdot x^{n-1} \]
This derivative exists for all \( x \) except when \( x = 0 \) if \( n - 1 < 0 \) (which implies \( n < 1 \)).

3. **Analyzing the Derivative**:
- If \( n < 1 \), then \( n - 1 < 0 \) and \( f'(x) \) becomes undefined at \( x = 0 \).
- If \( n = 1 \), then \( f'(x) = 1 \), which is defined for all \( x \).
- If \( n > 1 \), then \( n - 1 > 0 \) and \( f'(x) \) is defined for all \( x \).

4. **Conclusion**:
For \( f(x) = x^n \) to be differentiable for all \( x \), \( n \) must be greater than or equal to 1. Therefore, \( n \) can take any value in the interval:
\[ n \in [1, \infty) \]

### Final Answer:
The values of \( n \) for which the function \( f(x) = x^n \) is differentiable for all \( x \) are in the interval \( [1, \infty) \).

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