Let f(x) = x|x| then f'(0) is equal to

Options

A. 1
B. `-1`
C. 0
D. `+-1`

Answer

To find \( f'(0) \) for the function \( f(x) = x|x| \), we will follow these steps:

### Step 1: Define the function based on the value of \( x \)
The function \( f(x) = x|x| \) can be expressed differently based on whether \( x \) is positive or negative.

- For \( x \geq 0 \):
\[ f(x) = x \cdot x = x^2 \]

- For \( x < 0 \):
\[ f(x) = x \cdot (-x) = -x^2 \]

### Step 2: Check continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate \( f(0) \) and the limits from both sides.

- \( f(0) = 0|0| = 0 \)

Now, we check the left-hand limit and right-hand limit as \( x \) approaches 0:

- Right-hand limit (\( x \to 0^+ \)):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0 \]

- Left-hand limit (\( x \to 0^- \)):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x^2 = 0 \]

Since both limits equal \( f(0) \), the function is continuous at \( x = 0 \).

### Step 3: Differentiate the function
Next, we differentiate \( f(x) \) for both cases:

- For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx}(x^2) = 2x \]

- For \( x < 0 \):
\[ f'(x) = \frac{d}{dx}(-x^2) = -2x \]

### Step 4: Evaluate the derivative at \( x = 0 \)
Now we need to find \( f'(0) \). Since \( f'(x) \) is defined piecewise, we can use the right-hand derivative at \( x = 0 \):

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

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

Since both the left-hand and right-hand derivatives at \( x = 0 \) are equal, we conclude that:
\[ f'(0) = 0 \]

### Final Answer
Thus, \( f'(0) = 0 \).

---