Show that the function `f(x) = {{:(2x+3",",-3 le x lt -2),(x+1",",-2 le x lt 0),(x+2",",0 le x le 1):}` is discontinuous at x = 0 and continuous at every point in interval `[-3, 1]`

Answer

To show that the function

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

is discontinuous at \(x = 0\) and continuous at every other point in the interval \([-3, 1]\), we will check the continuity at the critical points: \(x = -3\), \(x = -2\), \(x = 0\), and \(x = 1\).

### Step 1: Check continuity at \(x = -3\)

1. **Right-hand limit**:
\[ \lim_{x \to -3^+} f(x) = \lim_{h \to 0} f(-3 + h) = \lim_{h \to 0} (2(-3 + h) + 3) = \lim_{h \to 0} (-6 + 2h + 3) = -3 \]

2. **Value of the function**:
\[ f(-3) = 2(-3) + 3 = -6 + 3 = -3 \]

Since the right-hand limit equals the value of the function:
\[ \lim_{x \to -3^+} f(x) = f(-3) = -3 \]
Thus, \(f(x)\) is continuous at \(x = -3\).

### Step 2: Check continuity at \(x = -2\)

1. **Left-hand limit**:
\[ \lim_{x \to -2^-} f(x) = \lim_{h \to 0} f(-2 - h) = \lim_{h \to 0} (2(-2 - h) + 3) = \lim_{h \to 0} (-4 - 2h + 3) = -1 \]

2. **Right-hand limit**:
\[ \lim_{x \to -2^+} f(x) = \lim_{h \to 0} f(-2 + h) = \lim_{h \to 0} (-2 + h + 1) = -1 \]

3. **Value of the function**:
\[ f(-2) = -2 + 1 = -1 \]

Since the left-hand limit, right-hand limit, and the value of the function are equal:
\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2) = -1 \]
Thus, \(f(x)\) is continuous at \(x = -2\).

### Step 3: Check continuity at \(x = 0\)

1. **Left-hand limit**:
\[ \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} (-h + 1) = 1 \]

2. **Right-hand limit**:
\[ \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} (h + 2) = 2 \]

Since the left-hand limit and right-hand limit are not equal:
\[ \lim_{x \to 0^-} f(x) = 1 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 2 \]
Thus, \(f(x)\) is discontinuous at \(x = 0\).

### Step 4: Check continuity at \(x = 1\)

1. **Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} (1 - h + 2) = 3 \]

2. **Value of the function**:
\[ f(1) = 1 + 2 = 3 \]

Since the left-hand limit equals the value of the function:
\[ \lim_{x \to 1^-} f(x) = f(1) = 3 \]
Thus, \(f(x)\) is continuous at \(x = 1\).

### Conclusion

The function \(f(x)\) is continuous at \(x = -3\), \(x = -2\), and \(x = 1\), but discontinuous at \(x = 0\). Therefore, we conclude that:

- \(f(x)\) is discontinuous at \(x = 0\).
- \(f(x)\) is continuous at every other point in the interval \([-3, 1]\).