If `f(x)={(-x^2",",when xle0),(5x-4",","when "0 lt x le 1),(4x^2-3x",","when "1 lt x lt 2),(3x+4",",when xge2):}` ,then

Options

A. f(x) is continuous at x = 0
B. f(x) is continuous at x = 2
C. f(x) is discontinuous at x = 1
D. None of these

Answer

To determine the points of continuity or discontinuity for the piecewise function

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

we will check the continuity at the points where the definition of the function changes, which are \(x = 0\), \(x = 1\), and \(x = 2\).

### Step 1: Check continuity at \(x = 0\)

1. **Find \(f(0)\)**:
\[ f(0) = -0^2 = 0 \]

2. **Find \(\lim_{x \to 0^-} f(x)\)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0 \]

3. **Find \(\lim_{x \to 0^+} f(x)\)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (5x - 4) = 5(0) - 4 = -4 \]

4. **Compare the limits and the function value**:
\[ \lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = -4, \quad f(0) = 0 \]
Since \(\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)\), the function is discontinuous at \(x = 0\).

### Step 2: Check continuity at \(x = 1\)

1. **Find \(f(1)\)**:
\[ f(1) = 5(1) - 4 = 1 \]

2. **Find \(\lim_{x \to 1^-} f(x)\)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5x - 4) = 5(1) - 4 = 1 \]

3. **Find \(\lim_{x \to 1^+} f(x)\)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^2 - 3x) = 4(1)^2 - 3(1) = 4 - 3 = 1 \]

4. **Compare the limits and the function value**:
\[ \lim_{x \to 1^-} f(x) = 1, \quad \lim_{x \to 1^+} f(x) = 1, \quad f(1) = 1 \]
Since all values are equal, the function is continuous at \(x = 1\).

### Step 3: Check continuity at \(x = 2\)

1. **Find \(f(2)\)**:
\[ f(2) = 3(2) + 4 = 6 + 4 = 10 \]

2. **Find \(\lim_{x \to 2^-} f(x)\)**:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x^2 - 3x) = 4(2)^2 - 3(2) = 16 - 6 = 10 \]

3. **Find \(\lim_{x \to 2^+} f(x)\)**:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + 4) = 3(2) + 4 = 6 + 4 = 10 \]

4. **Compare the limits and the function value**:
\[ \lim_{x \to 2^-} f(x) = 10, \quad \lim_{x \to 2^+} f(x) = 10, \quad f(2) = 10 \]
Since all values are equal, the function is continuous at \(x = 2\).

### Summary of Continuity
- The function is discontinuous at \(x = 0\).
- The function is continuous at \(x = 1\).
- The function is continuous at \(x = 2\).

### Final Answer
The function is discontinuous at \(x = 0\) and continuous at \(x = 1\) and \(x = 2\).