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To analyze this question, let’s go through each option one by one for the function \( f(x) = x^2 \sin(\sin(x^{-1})) \) with \( f(0) = 0 \).

1. **Option (A): Differentiable Everywhere**

   To check differentiability everywhere, we need to investigate the behavior of \( f(x) \) at \( x = 0 \) and for other values of \( x \neq 0 \).
   
   - For \( x \neq 0 \), the function \( f(x) = x^2 \sin(\sin(x^{-1})) \) is differentiable because it is the product of \( x^2 \) (a differentiable function) and \( \sin(\sin(x^{-1})) \) (bounded for all \( x \)).
   
   - At \( x = 0 \), we must verify the differentiability. A function is differentiable at \( x = 0 \) if the limit \( \lim_{x \to 0} \frac{f(x) - f(0)}{x} \) exists. Here,
     \[
     \frac{f(x) - f(0)}{x} = \frac{x^2 \sin(\sin(x^{-1}))}{x} = x \sin(\sin(x^{-1})).
     \]
     Since \( \sin(\sin(x^{-1})) \) oscillates between -1 and 1 as \( x \to 0 \), the limit does not converge to a single value. Therefore, \( f(x) \) is **not differentiable at \( x = 0 \)**.
   
   Hence, (A) is incorrect.

2. **Option (B): \( f'(0) \neq 0 \)**

   From the previous calculation, we saw that the derivative \( f'(0) \) does not exist because the limit \( \lim_{x \to 0} \frac{f(x) - f(0)}{x} \) does not converge. Therefore, \( f'(0) \) is undefined rather than nonzero. So, (B) is incorrect.

3. **Option (C): \( f'(0) = 0 \)**

   Since \( f'(0) \) does not exist, it cannot equal 0. Thus, (C) is also incorrect.

**Conclusion:** None of the options (A), (B), or (C) is correct because \( f(x) \) is not differentiable at \( x = 0 \) and \( f'(0) \) does not exist.

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### Related Questions
1. How is differentiability at a point evaluated for functions with oscillating terms?
2. What conditions ensure the differentiability of a piecewise-defined function?
3. Why does the oscillatory behavior of \( \sin(\sin(x^{-1})) \) affect differentiability at \( x = 0 \)?
4. How do bounded functions behave when multiplied by terms approaching zero?
5. What techniques are useful for proving non-differentiability at specific points?

### Tip
When analyzing differentiability at a specific point, checking the behavior of the function’s limit is crucial, especially with oscillatory or undefined terms.

Which of the following is true for the function f(x) = x^2 sin(sin(x^-1)) with f(0) = 0?

Explore the differentiability of the function f(x) = x^2 sin(sin(x^-1)) at x = 0. This problem addresses the effects of oscillatory terms on differentiability and uses limit-based analysis for solving.

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