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Magazine Chapter 7: Problem 5
For what positive values of \(p\) is the function \(f(x)=|x|^{p}\) differentiable at \(0 ?\)
Short Answer
Expert verified
The function is differentiable at 0 for \( p > 1 \).
Step by step solution
01
Define Differentiability at a Point
A function is differentiable at a point if the derivative exists at that point. For the function \( f(x) = |x|^p \), we need to check whether \( \lim_{{h \to 0}} \frac{f(h) - f(0)}{h} \) exists to determine if \( f(x) \) is differentiable at \( x = 0 \). Since \( f(0) = |0|^p = 0 \), the expression simplifies to \( \lim_{{h \to 0}} \frac{|h|^p}{h} \).
02
Consider Right and Left Limits
We need to compute two one-sided limits: \( \lim_{{h \to 0^+}} \frac{|h|^p}{h} \) and \( \lim_{{h \to 0^-}} \frac{|h|^p}{h} \). For \( h > 0 \), the absolute value \( |h| = h \), so \( \frac{|h|^p}{h} = h^{p-1} \). For \( h < 0 \), the absolute value \( |h| = -h \), giving \( \frac{(-h)^p}{h} = (-1)^p (-h)^{p-1} \).
03
Analyze the Right-Side Limit
Evaluate \( \lim_{{h \to 0^+}} h^{p-1} \). As \( h \to 0^+ \), if \( p-1 > 0 \) (i.e., \( p > 1 \)), this limit approaches 0. If \( p-1 = 0 \) (i.e., \( p = 1 \)), the limit is indeterminate (approaches 1). If \( p-1 < 0 \) (i.e., \( p < 1 \)), the limit does not exist. Thus, for differentiability, \( p > 1 \) must hold for the right limit to approach 0.
04
Analyze the Left-Side Limit
Evaluate \( \lim_{{h \to 0^-}} (-1)^p (-h)^{p-1} \). We need to consider the parity of \( p \). If \( p \) is an integer, then \((-1)^p = 1\) for even \(p\) and \((-1)\) for odd \(p\). When \( p - 1 > 0 \), \((-h)^{p-1} \to 0 \) as \( h \to 0^- \). Thus, for the left limit to approach 0, the condition \( p > 1 \) is again required.
05
Conclusion
Both one-sided limits must exist and match for the derivative to exist at \( x = 0 \). Analyzing both limits, we find that \( p > 1 \) ensures both limits exist and match to 0. Hence, the function \( f(x) = |x|^p \) is differentiable at \( x=0 \) for all \( p > 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiability
When we talk about differentiability, we refer to a function's ability to have a defined derivative at a certain point. A derivative is simply the rate at which a function changes at that point. For the function \[ f(x) = |x|^p \] to be differentiable at zero, the derivative must exist. This involves ensuring the limit \[ \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] exists. Since in this case, \[ f(0) = |0|^p = 0, \]the expression becomes \[ \lim_{h \to 0} \frac{|h|^p}{h}. \]If the right and left limits as \( h \) approaches zero are equal, the function is differentiable at that point. That's why we focus on one-sided limits.
Limits
Limits are a fundamental concept in calculus, serving as a building block for derivatives and integrals. In essence, a limit describes the behavior of a function as its input approaches a specific value. To determine if \[ f(x) = |x|^p \] is differentiable at zero, we need to explore the one-sided limits of \[ \frac{|h|^p}{h} \] as \( h \) approaches zero from the positive and negative sides.
- Right-Side Limit: As \( h \to 0^+ \), the expression becomes \( h^{p-1} \). For this limit to exist and be zero, \( p-1 \) must be positive, or \( p > 1 \).
- Left-Side Limit: As \( h \to 0^- \), consider the parity of \( p \). For even \( p \), \((-h)^p\) equals \( h^p \), and for odd \( p \), \((-h)^p\) equals \( -h^p \). The expression is {(-1)^p (-h)^{p-1}.\} To ensure the result is zero, again \( p > 1 \) is needed.
Positive Values of p
The exercise focuses on determining for which positive values of \( p \) the function \[ f(x) = |x|^p \] is differentiable at zero. The condition for differentiability, as seen from our examination of the limits, is that the power \( p \) must be greater than 1. Here鈥檚 why:
- When \( p > 1 \), both the right and left limits we discussed converge to zero, ensuring a smooth transition at \( x = 0 \).
- If \( p = 1 \), the limits become indeterminate. The expression doesn鈥檛 lead to a defined derivative at zero because it results in contrasting values rather than a specific number.
- When \( p < 1 \), both the right and left limits differ significantly, suggesting a discontinuous change at \( x = 0 \).
