Let `f(x) = {{:(-1,-2 le x lt 0),(x^2-1,0 le x le 2):}` and `g(x)=|f(x)|+f(|x|)`. Then , in the interval (-2,2),g is

Options

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

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) step by step.

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

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

### Step 2: Analyze the function \( f(x) \)

1. For \( -2 \leq x < 0 \), \( f(x) = -1 \). This is a constant function.
2. For \( 0 \leq x \leq 2 \), \( f(x) = x^2 - 1 \). This is a quadratic function, which is continuous and differentiable everywhere in its domain.

### Step 3: Identify points of non-differentiability in \( f(x) \)

The only potential point of non-differentiability is at \( x = 0 \), where the definition of \( f(x) \) changes. To check differentiability at this point, we calculate the left-hand and right-hand derivatives:

- Left-hand derivative at \( x = 0 \):
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-1 - (-1)}{h} = 0 \]

- Right-hand derivative at \( x = 0 \):
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(h^2 - 1) - (-1)}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = 0 \]

Since both derivatives are equal, \( f(x) \) is differentiable at \( x = 0 \).

### Step 4: Define the function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = |f(x)| + f(|x|) \]

### Step 5: Analyze \( |f(x)| \)

1. For \( -2 \leq x < 0 \), \( f(x) = -1 \) implies \( |f(x)| = 1 \).
2. For \( 0 \leq x \leq 2 \), \( f(x) = x^2 - 1 \) implies \( |f(x)| = 1 - x^2 \) (since \( x^2 - 1 \) is negative for \( 0 \leq x < 1 \) and positive for \( 1 \leq x \leq 2 \)).

### Step 6: Analyze \( f(|x|) \)

1. For \( -2 < x < 0 \), \( |x| = -x \) (which is positive), so \( f(|x|) = f(-x) = -1 \).
2. For \( 0 \leq x \leq 2 \), \( f(|x|) = f(x) = x^2 - 1 \).

### Step 7: Combine to find \( g(x) \)

Now we can write \( g(x) \) for different intervals:
- For \( -2 < x < 0 \):
\[ g(x) = 1 + (-1) = 0 \]
- For \( 0 \leq x < 1 \):
\[ g(x) = (1 - x^2) + (x^2 - 1) = 0 \]
- For \( 1 \leq x \leq 2 \):
\[ g(x) = (1 - x^2) + (x^2 - 1) = 0 \]

### Step 8: Identify points of non-differentiability in \( g(x) \)

The function \( g(x) \) is constant (0) in the intervals defined. The only point to check is at \( x = 0 \) where \( g(x) \) transitions from one definition to another.

### Conclusion

Since \( g(x) \) is constant in the intervals and only changes at \( x = 0 \), we conclude that \( g(x) \) is non-differentiable at one point (at \( x = 0 \)).

Thus, the answer is that \( g \) is non-differentiable at one point in the interval (-2, 2).