Consider `f(x)={{:(-2",",-1lexlt0),(x^(2)-2",",0lexle2):}` and `g(x)=|f(x)|+f(|x|)`. Then, in the interval `(-2, 2), g(x)` is

Options

A. not differentiable at one point
B. differentiable at all points
C. not continuous
D. not differentiable at two points

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) based on the given piecewise definition of \( f(x) \).

### Step 1: Define \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} -2 & \text{if } -2 < x < 0 \\ x^2 - 2 & \text{if } 0 \leq x \leq 2 \end{cases} \]

### Step 2: Calculate \( |f(x)| \)
Next, we need to find \( |f(x)| \):
- For \( -2 < x < 0 \): \( f(x) = -2 \) so \( |f(x)| = 2 \).
- For \( 0 \leq x \leq 2 \): \( f(x) = x^2 - 2 \). The value of \( |f(x)| \) will depend on whether \( x^2 - 2 \) is positive or negative:
- For \( 0 \leq x < \sqrt{2} \): \( f(x) < 0 \) so \( |f(x)| = 2 - x^2 \).
- For \( \sqrt{2} \leq x \leq 2 \): \( f(x) \geq 0 \) so \( |f(x)| = x^2 - 2 \).

### Step 3: Calculate \( f(|x|) \)
Now, we calculate \( f(|x|) \):
- For \( -2 < x < 0 \): \( |x| = -x \) which is in the range \( (0, 2) \), so \( f(|x|) = f(-x) = (-x)^2 - 2 = x^2 - 2 \).
- For \( 0 \leq x < 2 \): \( f(|x|) = f(x) = x^2 - 2 \).
- For \( x < -2 \): \( |x| > 2 \) which is outside the definition of \( f(x) \).

### Step 4: Define \( g(x) \)
Now we can define \( g(x) \):
\[ g(x) = |f(x)| + f(|x|) \]
We will analyze \( g(x) \) in the intervals \( (-2, 0) \), \( (0, \sqrt{2}) \), \( (\sqrt{2}, 2) \).

#### Interval 1: \( -2 < x < 0 \)
\[ g(x) = |f(x)| + f(|x|) = 2 + (x^2 - 2) = x^2 \]

#### Interval 2: \( 0 \leq x < \sqrt{2} \)
\[ g(x) = |f(x)| + f(|x|) = (2 - x^2) + (x^2 - 2) = 0 \]

#### Interval 3: \( \sqrt{2} \leq x \leq 2 \)
\[ g(x) = |f(x)| + f(|x|) = (x^2 - 2) + (x^2 - 2) = 2x^2 - 4 \]

### Step 5: Analyze differentiability
To check differentiability, we need to find the derivative \( g'(x) \) in each interval:
- For \( -2 < x < 0 \): \( g'(x) = 2x \).
- For \( 0 < x < \sqrt{2} \): \( g'(x) = 0 \).
- For \( \sqrt{2} < x < 2 \): \( g'(x) = 4x \).

Now, we check the points of interest:
- At \( x = 0 \): Left-hand derivative \( g'(-0) = 0 \) and right-hand derivative \( g'(0) = 0 \) → differentiable.
- At \( x = \sqrt{2} \): Left-hand derivative \( g'(\sqrt{2}^-) = 0 \) and right-hand derivative \( g'(\sqrt{2}^+) = 4\sqrt{2} \) → not differentiable.

### Conclusion
The function \( g(x) \) is differentiable everywhere in the interval \( (-2, 2) \) except at \( x = \sqrt{2} \).