Let `f(x)={{:(-2",",-3lexle0),(x-2",",0ltxle3):}andg(x)=f(|x|)+|f(x)|` Which of the following statement is correct ? g(x) is differentiable at x=0 g(x) is differentiable at x=2

Options

A. 1 olny
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To solve the problem, we need to analyze the function \( g(x) = f(|x|) + |f(x)| \) and check its differentiability at \( x = 0 \) and \( x = 2 \).

### Step 1: Define the function \( f(x) \)

The function \( f(x) \) is defined piecewise as follows:
\[ f(x) = \begin{cases} -2 & \text{for } -3 \leq x \leq 0 \\ x - 2 & \text{for } 0 < x \leq 3 \end{cases} \]

### Step 2: Analyze \( g(x) \)

The function \( g(x) \) is given by:
\[ g(x) = f(|x|) + |f(x)| \]

#### Case 1: \( x = 0 \)

1. **Calculate \( f(0) \)**:
\[ f(0) = -2 \]
Therefore, \( |f(0)| = 2 \).

2. **Calculate \( g(0) \)**:
\[ g(0) = f(0) + |f(0)| = -2 + 2 = 0 \]

3. **Left-hand derivative at \( x = 0 \)**:
\[ g(-h) = f(h) + |f(-h)| \quad \text{for small } h > 0 \]
For \( h \) approaching 0 from the left (\( -h \)):
- For \( -h \) in the interval \( [-3, 0] \), \( f(-h) = -2 \).
- Thus, \( g(-h) = -2 + 2 = 0 \).

The left-hand derivative is:
\[ \lim_{h \to 0^-} \frac{g(-h) - g(0)}{-h} = \lim_{h \to 0^-} \frac{0 - 0}{-h} = 0 \]

4. **Right-hand derivative at \( x = 0 \)**:
For \( h \) approaching 0 from the right (\( h \)):
- For \( h \) in the interval \( (0, 3] \), \( f(h) = h - 2 \).
- Thus, \( g(h) = (h - 2) + |h - 2| \).

- If \( h < 2 \), \( |h - 2| = 2 - h \):
\[ g(h) = (h - 2) + (2 - h) = 0 \]
- If \( h \geq 2 \), \( |h - 2| = h - 2 \):
\[ g(h) = (h - 2) + (h - 2) = 2h - 4 \]

Therefore, for \( h \) approaching 0 from the right:
\[ g(h) = 0 \quad \text{(since \( h < 2 \))} \]

The right-hand derivative is:
\[ \lim_{h \to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0 \]

5. **Conclusion at \( x = 0 \)**:
Since both left-hand and right-hand derivatives at \( x = 0 \) are equal (both are 0), \( g(x) \) is differentiable at \( x = 0 \).

### Step 3: Analyze \( g(x) \) at \( x = 2 \)

1. **Calculate \( g(2) \)**:
\[ f(2) = 2 - 2 = 0 \quad \text{and thus } |f(2)| = 0 \]
\[ g(2) = f(2) + |f(2)| = 0 + 0 = 0 \]

2. **Left-hand derivative at \( x = 2 \)**:
For \( h \) approaching 0 from the left:
- For \( 2 - h \) (where \( h \) is small and positive), \( f(2 - h) = (2 - h) - 2 = -h \).
- Thus, \( |f(2 - h)| = h \).
\[ g(2 - h) = (-h) + h = 0 \]
The left-hand derivative is:
\[ \lim_{h \to 0^-} \frac{g(2 - h) - g(2)}{-h} = \lim_{h \to 0^-} \frac{0 - 0}{-h} = 0 \]

3. **Right-hand derivative at \( x = 2 \)**:
For \( h \) approaching 0 from the right:
- For \( 2 + h \), \( f(2 + h) = 2 + h - 2 = h \).
- Thus, \( |f(2 + h)| = h \).
\[ g(2 + h) = h + h = 2h \]
The right-hand derivative is:
\[ \lim_{h \to 0^+} \frac{g(2 + h) - g(2)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2 \]

4. **Conclusion at \( x = 2 \)**:
The left-hand derivative is 0, and the right-hand derivative is 2. Since they are not equal, \( g(x) \) is not differentiable at \( x = 2 \).

### Final Conclusion

- \( g(x) \) is differentiable at \( x = 0 \).
- \( g(x) \) is not differentiable at \( x = 2 \).

Thus, the correct statement is:
- **g(x) is differentiable at x=0.**