The points of discontinuity of the function
`f (x) = {{:(3x + 1"," ,0 le x lt2),(4x - 1"," ,2 lt x le 6),(5x + 2",", 6 lt x le 10 ):}`are:

Options

A. 6
B. 2
C. 6, 2
D. 6, 2, 0

Answer

To find the points of discontinuity of the piecewise function

\[ f(x) = \begin{cases} 3x + 1 & \text{if } 0 \leq x < 2 \\ 4x - 1 & \text{if } 2 < x \leq 6 \\ 5x + 2 & \text{if } 6 < x \leq 10 \end{cases} \]

we need to check the points where the definition of the function changes, which are at \(x = 2\) and \(x = 6\).

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

1. **Left-hand limit as \(x\) approaches 2**:
\[ \lim_{x \to 2^-} f(x) = 3(2) + 1 = 6 + 1 = 7 \]

2. **Right-hand limit as \(x\) approaches 2**:
\[ \lim_{x \to 2^+} f(x) = 4(2) - 1 = 8 - 1 = 7 \]

3. **Value of the function at \(x = 2\)**:
\[ f(2) = 4(2) - 1 = 8 - 1 = 7 \]

Since both the left-hand limit and right-hand limit at \(x = 2\) are equal to the function value, \(f(2)\), the function is continuous at \(x = 2\).

### Step 2: Check continuity at \(x = 6\)

1. **Left-hand limit as \(x\) approaches 6**:
\[ \lim_{x \to 6^-} f(x) = 4(6) - 1 = 24 - 1 = 23 \]

2. **Right-hand limit as \(x\) approaches 6**:
\[ \lim_{x \to 6^+} f(x) = 5(6) + 2 = 30 + 2 = 32 \]

3. **Value of the function at \(x = 6\)**:
\[ f(6) = 5(6) + 2 = 30 + 2 = 32 \]

Since the left-hand limit (23) does not equal the right-hand limit (32) and does not equal the function value \(f(6)\), the function is discontinuous at \(x = 6\).

### Conclusion

The only point of discontinuity for the function \(f(x)\) is at \(x = 6\).

### Final Answer
The points of discontinuity of the function are: \(x = 6\).

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