Suppose the function `f(x)=x^(n), n ne0` 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

Answer

To solve the problem, we need to analyze the differentiability of the function \( f(x) = x^n \) where \( n \neq 0 \).

### Step-by-step Solution:

1. **Identify the Function**:
We start with the function given in the problem:
\[ f(x) = x^n \]
where \( n \neq 0 \).

2. **Differentiate the Function**:
We will differentiate \( f(x) \) with respect to \( x \) using the power rule of differentiation:
\[ f'(x) = \frac{d}{dx}(x^n) = n x^{n-1} \]

3. **Determine Conditions for Differentiability**:
For \( f(x) \) to be differentiable for all \( x \), the derivative \( f'(x) \) must be defined for all \( x \). The term \( x^{n-1} \) is defined for all \( x \) except when \( x = 0 \) if \( n-1 < 0 \) (which implies \( n < 1 \)). Therefore, we need to ensure that \( n - 1 \geq 0 \) for \( f'(x) \) to be defined at \( x = 0 \).

4. **Set Up the Inequality**:
From the condition derived, we have:
\[ n - 1 \geq 0 \]
This simplifies to:
\[ n \geq 1 \]

5. **Conclusion**:
Thus, the interval for \( n \) is:
\[ n \in [1, \infty) \]

### Final Answer:
The values of \( n \) can be any element in the interval \( [1, \infty) \).

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