Discuss the continutiy of the function `f(x)= {(4x-2", " x le 2),(3x", " x gt 2):}`

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} 4x - 2 & \text{if } x \leq 2 \\ 3x & \text{if } x > 2 \end{cases} \]

at \( x = 2 \), we need to check the following conditions:

1. \( f(2) \) exists.
2. The left-hand limit \( \lim_{x \to 2^-} f(x) \) exists.
3. The right-hand limit \( \lim_{x \to 2^+} f(x) \) exists.
4. The left-hand limit and right-hand limit must be equal to \( f(2) \).

### Step 1: Calculate \( f(2) \)

Since \( x = 2 \) falls under the first case of the function definition:

\[ f(2) = 4(2) - 2 = 8 - 2 = 6 \]

### Step 2: Calculate the left-hand limit \( \lim_{x \to 2^-} f(x) \)

For \( x < 2 \), we use the first case of the function:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x - 2) \]

Substituting \( x = 2 \):

\[ = 4(2) - 2 = 8 - 2 = 6 \]

### Step 3: Calculate the right-hand limit \( \lim_{x \to 2^+} f(x) \)

For \( x > 2 \), we use the second case of the function:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x) \]

Substituting \( x = 2 \):

\[ = 3(2) = 6 \]

### Step 4: Check continuity

Now we have:

- \( f(2) = 6 \)
- \( \lim_{x \to 2^-} f(x) = 6 \)
- \( \lim_{x \to 2^+} f(x) = 6 \)

Since all three values are equal:

\[ f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 6 \]

### Conclusion

The function \( f(x) \) is continuous at \( x = 2 \).

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