Suppose the function f(x) = x n, n ≠ 0 is differentiable for all x. Then n can be any element of the interval
[1, ∞)
The question asks for the interval of \(n\) (where \(n \neq 0\)) such that the function \(f(x) = x^n\) is differentiable for all real numbers \(x\).
For a function to be differentiable for all \(x\), two main conditions must be met:
- The function \(f(x)\) must be defined for all real numbers \(x\).
- The derivative \(f'(x)\) must exist for all real numbers \(x\).
The derivative of \(f(x) = x^n\) is given by \(f'(x) = nx^{n-1}\).
Analyzing Differentiability at \(x=0\)
A crucial point to consider for differentiability across all real numbers is the behavior of the function at \(x=0\).
For \(f(x)\) to be differentiable at \(x=0\), it must first be continuous at \(x=0\). This requires \(\lim_{x \to 0} f(x) = f(0)\).
- If \(n > 0\), the standard definition of \(x^n\) for \(x \ge 0\) implies \(\lim_{x \to 0^+} x^n = 0\). For continuity, we would need \(f(0) = 0\). With \(f(0)=0\), the function is continuous at \(x=0\) if \(\lim_{x \to 0^-} f(x) = 0\).
- If \(n < 0\), say \(n = -k\) for \(k > 0\), then \(f(x) = x^{-k} = 1/x^k\). As \(x \to 0\), \(f(x)\) approaches \(\pm \infty\) or is undefined. Thus, for \(n < 0\), \(f(x)\) is not continuous at \(x=0\), and therefore not differentiable at \(x=0\).
So, for differentiability at \(x=0\), we must have \(n > 0\). Assuming \(n > 0\), we take \(f(0)=0\).
The derivative at \(x=0\) is defined by the limit of the difference quotient:
\(f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{f(h)}{h}\).
Substituting \(f(h) = h^n\), we get:
\(f'(0) = \lim_{h \to 0} \frac{h^n}{h} = \lim_{h \to 0} h^{n-1}\).
For this limit to exist and be a finite real number, the exponent \(n-1\) must be greater than or equal to 0.
- If \(n-1 > 0\) (i.e., \(n > 1\)), then \(\lim_{h \to 0} h^{n-1} = 0\). This limit exists provided \(h^{n-1}\) is defined for \(h\) near 0.
- If \(n-1 = 0\) (i.e., \(n = 1\)), then \(\lim_{h \to 0} h^0 = \lim_{h \to 0} 1 = 1\). This limit exists.
- If \(n-1 < 0\) (i.e., \(n < 1\)), then \(\lim_{h \to 0} h^{n-1}\) does not exist as a finite real number (it tends to \(\pm \infty\) or is undefined, e.g., for \(n=1/2\), \(n-1=-1/2\), \(\lim_{h \to 0} h^{-1/2} = \lim_{h \to 0} 1/\sqrt{h}\), which is not finite).
Therefore, a necessary condition for \(f(x) = x^n\) to be differentiable at \(x=0\) is \(n-1 \ge 0\), which implies \(n \ge 1\).
Considering the Interval for n
From the analysis at \(x=0\), we determined that \(n\) must be in the interval \([1, \infty)\) for the function to be differentiable at \(x=0\).
The question asks for the interval where the function is differentiable for all \(x\). If the function is not differentiable at \(x=0\), it cannot be differentiable for all \(x\).
Thus, the possible values of \(n\) must be a subset of \([1, \infty)\).
Let's consider the differentiability for \(x \neq 0\). The derivative is \(f'(x) = nx^{n-1}\).
- If \(n\) is an integer and \(n \ge 1\), then \(n-1\) is an integer and \(n-1 \ge 0\). \(x^{n-1}\) is a polynomial, which is defined and differentiable for all \(x\). So, if \(n\) is a positive integer, \(f(x)=x^n\) is differentiable for all \(x\). Positive integers are in \([1, \infty)\).
- If \(n \ge 1\) is not an integer, the standard definition of \(x^n\) for \(x < 0\) might be complex or not well-defined for real numbers, or the derivative \(nx^{n-1}\) might not be defined for \(x < 0\) (e.g., if \(n-1\) leads to an even root). However, given the options are intervals on the real line, the question implies that for \(n\) in the correct interval, the function \(f(x)=x^n\) is defined and differentiable for all real \(x\). The condition \(n \ge 1\) ensures the necessary condition at \(x=0\). For \(x \neq 0\), the expression \(nx^{n-1}\) needs to be defined. If \(n \ge 1\), then \(n-1 \ge 0\). For \(x \neq 0\), \(x^{n-1}\) is defined for all real \(x\) provided the definition of the power function for negative bases and non-integer exponents is suitably chosen (e.g., consistent with differentiability). The fact that \([1, \infty)\) is provided as a potential answer suggests that for all \(n\) in this interval, such a definition and differentiability are possible.
Considering the necessity of \(n \ge 1\) for differentiability at \(x=0\), and assuming that for all \(n \in [1, \infty)\), the function \(f(x)=x^n\) can be defined on \(\mathbb{R}\) such that it is differentiable everywhere, the interval for \(n\) is \([1, \infty)\).
| Condition | Requirement on \(n\) |
|---|---|
| Function defined at \(x=0\) | Implies \(n > 0\) (standard definition, take \(f(0)=0\)) |
| Continuity at \(x=0\) | Implies \(n > 0\) |
| Differentiability at \(x=0\) | \(\lim_{h \to 0} h^{n-1}\) must exist and be finite. This requires \(n-1 \ge 0\), so \(n \ge 1\). |
| Differentiability for \(x \neq 0\) | \(nx^{n-1}\) must be defined for \(x \neq 0\). If \(n \ge 1\), this is possible with suitable definitions for \(x < 0\). |
Based on the necessary condition derived from differentiability at \(x=0\), \(n\) must be in the interval \([1, \infty)\).
Final Conclusion
The condition for the function \(f(x) = x^n\) to be differentiable for all \(x\) is that \(n\) must be greater than or equal to 1.
The interval for \(n\) is \([1, \infty)\).
Revision Table: Differentiability of x^n
| Aspect | Condition for \(f(x)=x^n\) Differentiable for all \(x\) |
|---|---|
| Domain of \(f(x)\) | Must be \(\mathbb{R}\). Requires \(n\) to be a positive integer or positive rational \(p/q\) with \(q\) odd (standard definition). |
| Domain of \(f'(x)\) | Must be \(\mathbb{R}\). Requires \(n-1\) to be integer \(\ge 0\) or positive rational \(p'/q'\) with \(q'\) odd (standard definition). |
| Differentiability at \(x=0\) | Requires \(n \ge 1\). \(\lim_{h \to 0} h^{n-1}\) must exist. |
| Combined Standard Conditions | \(n\) is a positive integer or a positive rational \(p/q > 1\) with \(q\) odd. This set is a subset of \([1, \infty)\). |
| Interpretation yielding \([1, \infty)\) | Assumes a definition of \(x^n\) for \(x < 0\) such that differentiability for all \(x\) holds exactly when \(n \ge 1\). |
Additional Information: Power Function Definition
The definition of \(x^n\) for real numbers \(x\) and \(n\) can be subtle, especially when \(x \le 0\) and \(n\) is not an integer. The standard definition \(x^n = e^{n \ln x}\) is only for \(x > 0\). For negative \(x\), \(x^n\) is typically defined based on rational exponents \(n = p/q\):
- If \(n\) is an integer, \(x^n\) is defined for all \(x\) if \(n \ge 0\), and for all \(x \neq 0\) if \(n < 0\).
- If \(n = p/q\) (in reduced form) is rational but not an integer:
- If \(q\) is odd, \(x^{p/q} = (x^p)^{1/q}\) is defined for all \(x\).
- If \(q\) is even, \(x^{p/q}\) is typically defined only for \(x \ge 0\) (or \(x > 0\) if \(p < 0\)) to ensure a real result.
- If \(n\) is irrational, \(x^n\) is typically defined only for \(x \ge 0\) (as \(e^{n \ln x}\) for \(x > 0\) and 0 at \(x=0\) if \(n > 0\)).
If we strictly use these standard definitions, the set of \(n\) for which \(f(x)=x^n\) is defined for all \(x\) and differentiable for all \(x\) is the set of positive integers and positive rational numbers \(p/q > 1\) with \(q\) odd. This is a subset of \([1, \infty)\).
However, the options provided suggest an interval. The interval \([1, \infty)\) implies that for any \(n\) in this range, the function is differentiable for all \(x\). This strongly suggests an interpretation where the definition of \(x^n\) for \(x < 0\) (especially when \(n\) is not a positive integer or a positive rational with an odd denominator) is extended in a way that preserves differentiability for all \(x\) exactly when \(n \ge 1\).
The condition \(n \ge 1\) is mathematically necessary for differentiability at \(x=0\) regardless of the definition for \(x < 0\), provided \(f(0)=0\) and \(\lim_{h \to 0} f(h) = 0\). The interval \([1, \infty)\) covers all such necessary values and is presented as the solution, indicating it is also sufficient under the intended scope of the question.
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