If `f(x)={{:(,4,-3lt x lt -1),(,5+x,-1le x lt 0),(,5-x,0 le x lt 2),(,x^(2)+x-3,2 lt x lt 3):}` then, f(|x|) is

Options

A. differentiable but not continuous in (-3,3)
B. continuous but not differentiable in (-3,3)
C. continuous as well as differentiable in (-3,3)
D. neither continuous nor differentiable (-3,3)

Explanation

To solve the problem, we need to find the expression for \( f(|x|) \) based on the piecewise function defined for \( f(x) \).

Given the function:
\[ f(x) = \begin{cases} 4 & \text{for } -3 < x < -1 \\ 5 + x & \text{for } -1 \leq x < 0 \\ 5 - x & \text{for } 0 \leq x < 2 \\ x^2 + x - 3 & \text{for } 2 < x < 3 \end{cases} \]

### Step 1: Determine the ranges for \( |x| \)

The absolute value function \( |x| \) affects the input to \( f(x) \). We need to consider the cases for \( x \):

1. When \( x \geq 0 \), \( |x| = x \).
2. When \( x < 0 \), \( |x| = -x \).

### Step 2: Evaluate \( f(|x|) \) for different ranges of \( x \)

**Case 1: \( x \geq 0 \) (i.e., \( |x| = x \))**

- For \( 0 \leq x < 2 \):
\[ f(|x|) = f(x) = 5 - x \]

- For \( 2 \leq x < 3 \):
\[ f(|x|) = f(x) = x^2 + x - 3 \]

**Case 2: \( x < 0 \) (i.e., \( |x| = -x \))**

- For \( -3 < x < -2 \):
\[ f(|x|) = f(-x) = f(-(-x)) = f(x) = 4 \]

- For \( -2 \leq x < -1 \):
\[ f(|x|) = f(-x) = f(-(-x)) = f(x) = 5 + x \]

- For \( -1 \leq x < 0 \):
\[ f(|x|) = f(-x) = f(-(-x)) = f(x) = 5 + x \]

### Step 3: Combine the results

Now, we can combine the results from both cases to write the piecewise function for \( f(|x|) \):

\[ f(|x|) = \begin{cases} 4 & \text{for } -3 < x < -2 \\ 5 + x & \text{for } -2 \leq x < -1 \\ 5 - x & \text{for } 0 \leq x < 2 \\ x^2 + x - 3 & \text{for } 2 \leq x < 3 \end{cases} \]

### Final Answer

Thus, the function \( f(|x|) \) can be expressed as:
\[ f(|x|) = \begin{cases} 4 & \text{for } -3 < x < -2 \\ 5 + x & \text{for } -2 \leq x < 0 \\ 5 - x & \text{for } 0 \leq x < 2 \\ x^2 + x - 3 & \text{for } 2 \leq x < 3 \end{cases} \]