Let `f (x)= {{:(2-x"," , -3 le x le 0),( x-2"," , 0 lt x lt 4):}` Then `f ^(-1)` (x) is discontinous at x=

Answer

To determine where the inverse function \( f^{-1}(x) \) is discontinuous, we first need to find the inverse of the given piecewise function \( f(x) \).

### Step 1: Define the function \( f(x) \)
The function is defined as:
\[ f(x) = \begin{cases} 2 - x & \text{for } -3 \leq x \leq 0 \\ x - 2 & \text{for } 0 < x < 4 \end{cases} \]

### Step 2: Find the inverse for each piece of \( f(x) \)

#### For the first piece \( f(x) = 2 - x \) where \( -3 \leq x \leq 0 \):
To find the inverse, we set \( y = 2 - x \) and solve for \( x \):
\[ x = 2 - y \]
Thus, the inverse function for this interval is:
\[ f^{-1}(x) = 2 - x \quad \text{for } 2 \leq x \leq 5 \]
(Here, we find the range by substituting the endpoints of the domain: \( f(-3) = 5 \) and \( f(0) = 2 \)).

#### For the second piece \( f(x) = x - 2 \) where \( 0 < x < 4 \):
Setting \( y = x - 2 \) and solving for \( x \):
\[ x = y + 2 \]
Thus, the inverse function for this interval is:
\[ f^{-1}(x) = x + 2 \quad \text{for } -2 < x < 2 \]
(Here, we find the range by substituting the endpoints of the domain: \( f(0) = -2 \) and \( f(4) = 2 \)).

### Step 3: Combine the inverses
Now we can write the complete inverse function:
\[ f^{-1}(x) = \begin{cases} 2 - x & \text{for } 2 \leq x \leq 5 \\ x + 2 & \text{for } -2 < x < 2 \end{cases} \]

### Step 4: Check for continuity at the point where the pieces meet
The two pieces of the inverse function meet at \( x = 2 \). We need to check the continuity at this point.

#### Left-hand limit (LHL) as \( x \) approaches 2:
Using the second piece:
\[ \lim_{x \to 2^-} f^{-1}(x) = 2 + 2 = 4 \]

#### Right-hand limit (RHL) as \( x \) approaches 2:
Using the first piece:
\[ \lim_{x \to 2^+} f^{-1}(x) = 2 - 2 = 0 \]

### Step 5: Compare the limits
Since:
\[ \text{LHL} = 4 \quad \text{and} \quad \text{RHL} = 0 \]
we find that \( \text{LHL} \neq \text{RHL} \).

### Conclusion
Thus, the function \( f^{-1}(x) \) is discontinuous at:
\[ \boxed{2} \]