The set of allpoints of differentiability of the function `f(x) ={{:(x^(2) sin(1//x), x ne 0),(0, x =0):}` is

Options

A. `(-infty, 0)`
B. `(-infty, infty) ~ {0}`
C. `[0,infty)`
D. `(-infty, infty)`

Answer

To determine the set of all points of differentiability of the function

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we need to analyze the differentiability of \( f(x) \) at \( x = 0 \) and for \( x \neq 0 \).

### Step 1: Check differentiability at \( x \neq 0 \)

For \( x \neq 0 \), the function \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) is a product of two differentiable functions: \( x^2 \) and \( \sin\left(\frac{1}{x}\right) \). Since both functions are differentiable for \( x \neq 0 \), we conclude that \( f(x) \) is differentiable for all \( x \neq 0 \).

**Hint:** Check the differentiability of the function at points where it is defined.

### Step 2: Check differentiability at \( x = 0 \)

To check the differentiability at \( x = 0 \), we need to compute the left-hand and right-hand derivatives.

**Left-hand derivative:**

The left-hand derivative at \( x = 0 \) is given by:

\[ f'(0) = \lim_{h \to 0^-} \frac{f(0) - f(-h)}{-h} \]

Substituting \( f(0) = 0 \) and \( f(-h) = (-h)^2 \sin\left(\frac{1}{-h}\right) = h^2 \sin\left(-\frac{1}{h}\right) \):

\[ f'(0) = \lim_{h \to 0^-} \frac{0 - h^2 \sin\left(-\frac{1}{h}\right)}{-h} = \lim_{h \to 0^-} \frac{h^2 \sin\left(-\frac{1}{h}\right)}{h} \]

This simplifies to:

\[ f'(0) = \lim_{h \to 0^-} h \sin\left(-\frac{1}{h}\right) \]

Since \( \sin\left(-\frac{1}{h}\right) \) is bounded between -1 and 1, we have:

\[ -h \leq h \sin\left(-\frac{1}{h}\right) \leq h \]

As \( h \to 0 \), both bounds approach 0, so by the Squeeze Theorem:

\[ f'(0) = 0 \]

**Right-hand derivative:**

The right-hand derivative at \( x = 0 \) is given by:

\[ f'(0) = \lim_{h \to 0^+} \frac{f(0) - f(h)}{-h} \]

Substituting \( f(0) = 0 \) and \( f(h) = h^2 \sin\left(\frac{1}{h}\right) \):

\[ f'(0) = \lim_{h \to 0^+} \frac{0 - h^2 \sin\left(\frac{1}{h}\right)}{-h} = \lim_{h \to 0^+} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} \]

This simplifies to:

\[ f'(0) = \lim_{h \to 0^+} h \sin\left(\frac{1}{h}\right) \]

Again, since \( \sin\left(\frac{1}{h}\right) \) is bounded between -1 and 1, we have:

\[ -h \leq h \sin\left(\frac{1}{h}\right) \leq h \]

As \( h \to 0 \), both bounds approach 0, so by the Squeeze Theorem:

\[ f'(0) = 0 \]

### Conclusion

Since the left-hand derivative and right-hand derivative at \( x = 0 \) are both equal to 0, we conclude that \( f(x) \) is differentiable at \( x = 0 \).

Thus, the function \( f(x) \) is differentiable for all \( x \in \mathbb{R} \).

The set of all points of differentiability of the function is:

\[ (-\infty, \infty) \]