Consider the function `f(x)={:{(x^(2)|x|x!=0),(" 0 "x=0):}}` what is f'(0) equal to ?

Options

A. 0
B. 1
C. `-1`
D. It does not exist

Answer

To find the derivative \( f'(0) \) of the given piecewise function

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

we need to check if the function is differentiable at \( x = 0 \). This requires us to calculate the left-hand derivative (LHD) and the right-hand derivative (RHD) at that point.

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

For \( x \neq 0 \), we can express \( f(x) \) as:

- For \( x > 0 \):
\[ f(x) = x^2 \cdot x = x^3 \]

- For \( x < 0 \):
\[ f(x) = x^2 \cdot (-x) = -x^3 \]

### Step 2: Calculate the left-hand derivative (LHD) at \( x = 0 \)

The left-hand derivative is calculated using the limit:

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

Since \( h < 0 \), we use the expression for \( f(h) \):

\[ f(h) = -h^3 \]

Thus, we have:

\[ f'_{L}(0) = \lim_{h \to 0^-} \frac{-h^3}{h} = \lim_{h \to 0^-} -h^2 = 0 \]

### Step 3: Calculate the right-hand derivative (RHD) at \( x = 0 \)

The right-hand derivative is calculated similarly:

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

Since \( h > 0 \), we use the expression for \( f(h) \):

\[ f(h) = h^3 \]

Thus, we have:

\[ f'_{R}(0) = \lim_{h \to 0^+} \frac{h^3}{h} = \lim_{h \to 0^+} h^2 = 0 \]

### Step 4: Conclusion

Since both the left-hand derivative and the right-hand derivative at \( x = 0 \) are equal:

\[ f'_{L}(0) = f'_{R}(0) = 0 \]

We conclude that:

\[ f'(0) = 0 \]

### Final Answer

Thus, \( f'(0) = 0 \).

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