Consider, `f(x) = {{:(x^(2)/(|x|), x ne 0),(0, x =0):}`

Options

A. f(x) is discontinuous everywhere
B. f(x) is continuous everywhere
C. f'(x) exists in (-1,1}
D. f'(x) exists in (- 2,2)

Answer

To solve the problem, we need to analyze the function given by:

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

### Step 1: Simplify the function for \(x \neq 0\)

For \(x > 0\), \(|x| = x\), so:

\[ f(x) = \frac{x^2}{x} = x \quad \text{for } x > 0 \]

For \(x < 0\), \(|x| = -x\), so:

\[ f(x) = \frac{x^2}{-x} = -x \quad \text{for } x < 0 \]

Thus, we can rewrite the function as:

\[ f(x) = \begin{cases} x & \text{if } x > 0 \\ -x & \text{if } x < 0 \\ 0 & \text{if } x = 0 \end{cases} \]

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

To check if \(f(x)\) is continuous at \(x = 0\), we need to find the left-hand limit and the right-hand limit as \(x\) approaches \(0\).

**Left-hand limit** as \(x \to 0^{-}\):

\[ \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (-x) = 0 \]

**Right-hand limit** as \(x \to 0^{+}\):

\[ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x = 0 \]

Since both limits are equal to \(f(0)\):

\[ f(0) = 0 \]

Thus, \(f(x)\) is continuous at \(x = 0\).

### Step 3: Check continuity everywhere

Since \(f(x)\) is defined piecewise and both pieces are continuous for \(x > 0\) and \(x < 0\), we conclude that \(f(x)\) is continuous everywhere.

### Step 4: Check differentiability at \(x = 0\)

Now we need to check if \(f(x)\) is differentiable at \(x = 0\).

**Left-hand derivative** at \(x = 0\):

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

**Right-hand derivative** at \(x = 0\):

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

Since the left-hand derivative \(-1\) and the right-hand derivative \(1\) are not equal, \(f(x)\) is not differentiable at \(x = 0\).

### Conclusion

1. **Continuity**: The function \(f(x)\) is continuous everywhere.
2. **Differentiability**: The function \(f(x)\) is not differentiable at \(x = 0\).

### Final Answer

- \(f(x)\) is continuous everywhere.
- \(f(x)\) is not differentiable at \(x = 0\).

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