The set of points where the function `f(x) = x |x|` is differentiable is

Options

A. `(-infty, infty)`
B. `(-infty, 0) cup [0,infty)`
C. `(0,infty)`
D. `[0,infty)`

Answer

To determine the set of points where the function \( f(x) = x |x| \) is differentiable, we will follow these steps:

### Step 1: Rewrite the function using piecewise definition
The function \( f(x) = x |x| \) can be expressed in a piecewise manner based on the definition of the absolute value:

\[ f(x) = \begin{cases} -x^2 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} \]

### Step 2: Check differentiability at \( x = 0 \)
To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative at this point.

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

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

Since \( 0 + h \) is less than 0, we use the first case of the piecewise function:

\[ f(0 + h) = f(h) = -h^2 \quad \text{and} \quad f(0) = 0 \]

Thus, we have:

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

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

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

Since \( 0 + h \) is greater than 0, we use the second case of the piecewise function:

\[ f(0 + h) = f(h) = h^2 \quad \text{and} \quad f(0) = 0 \]

Thus, we have:

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

### Step 3: Compare the left-hand and right-hand derivatives
Since both the left-hand derivative and the right-hand derivative at \( x = 0 \) are equal:

\[ f'(0^-) = f'(0^+) = 0 \]

This means that \( f(x) \) is differentiable at \( x = 0 \).

### Step 4: Check differentiability for \( x < 0 \) and \( x > 0 \)
For \( x < 0 \), \( f(x) = -x^2 \) is a polynomial function, which is differentiable everywhere in this interval.

For \( x > 0 \), \( f(x) = x^2 \) is also a polynomial function, which is differentiable everywhere in this interval.

### Conclusion
Since \( f(x) \) is differentiable for all \( x < 0 \), at \( x = 0 \), and for all \( x > 0 \), we conclude that:

\[ \text{The function } f(x) = x |x| \text{ is differentiable for all } x \in (-\infty, \infty). \]