Math Problem Statement
Check the differentiability of the function f(x) = { x^2 * cos(1/x) if x ≠ 0, 0 if x = 0 } at x = 0 and x = 1/π.
Solution
The problem asks to check the differentiability of the function given by:
\begin{cases} x^2 \cos\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$$ at $$x = 0$$ and $$x = \frac{1}{\pi}$$. ### Step-by-Step Approach: #### 1. Differentiability at $$x = 0$$ To check differentiability at $$x = 0$$, we need to check: - Continuity at $$x = 0$$ - Existence of the derivative $$f'(0)$$ **Continuity at $$x = 0$$:** We check whether $$\lim_{x \to 0} f(x) = f(0)$$. Given $$f(0) = 0$$, we now find: $$\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)$$ Since $$\cos\left(\frac{1}{x}\right)$$ oscillates between -1 and 1 for all $$x$$, and since $$x^2$$ approaches 0 as $$x \to 0$$, we have: $$- x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2$$ By the squeeze theorem: $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0$$ Thus, $$\lim_{x \to 0} f(x) = 0 = f(0)$$, so $$f(x)$$ is continuous at $$x = 0$$. **Existence of derivative at $$x = 0$$:** We need to check if the derivative $$f'(0)$$ exists, defined as: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$$ Substitute $$f(0) = 0$$: $$f'(0) = \lim_{h \to 0} \frac{h^2 \cos\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \cos\left(\frac{1}{h}\right)$$ Since $$\cos\left(\frac{1}{h}\right)$$ oscillates between -1 and 1, and $$h \to 0$$, by the squeeze theorem: $$\lim_{h \to 0} h \cos\left(\frac{1}{h}\right) = 0$$ Thus, $$f'(0) = 0$$, and $$f(x)$$ is differentiable at $$x = 0$$. #### 2. Differentiability at $$x = \frac{1}{\pi}$$ Since $$x \neq 0$$, we can differentiate the function directly. For $$f(x) = x^2 \cos\left(\frac{1}{x}\right)$$, we apply the product rule: $$f'(x) = \frac{d}{dx}\left(x^2 \cos\left(\frac{1}{x}\right)\right) = 2x \cos\left(\frac{1}{x}\right) - \sin\left(\frac{1}{x}\right)$$ Now, substitute $$x = \frac{1}{\pi}$$: $$f'\left(\frac{1}{\pi}\right) = 2 \left(\frac{1}{\pi}\right) \cos(\pi) - \sin(\pi)$$ Since $$\cos(\pi) = -1$$ and $$\sin(\pi) = 0$$, we have: $$f'\left(\frac{1}{\pi}\right) = 2 \left(\frac{1}{\pi}\right)(-1) = -\frac{2}{\pi}$$ Thus, $$f(x)$$ is differentiable at $$x = \frac{1}{\pi}$$, and the derivative is $$f'\left(\frac{1}{\pi}\right) = -\frac{2}{\pi}$$. --- Let me know if you'd like more details! Here are 5 related questions to further explore the concepts: 1. What are the conditions for differentiability of a piecewise function? 2. How does the squeeze theorem work in proving limits? 3. Can a function be continuous but not differentiable at a point? 4. What role do oscillating functions like $$\cos\left(\frac{1}{x}\right)$$ play in limiting behavior? 5. How do product and chain rules combine when differentiating products of functions? **Tip:** When checking differentiability, always ensure that both continuity and the existence of the derivative are considered.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differentiability
Limits
Continuity
Piecewise Functions
Formulas
Squeeze Theorem
Definition of Differentiability: f'(x) = lim(h→0) [f(x+h) - f(x)]/h
Theorems
Squeeze Theorem
Continuity and Differentiability Theorems
Suitable Grade Level
Undergraduate
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