Discuss the continutiy of the function
`f(x)= {(3x+5", " x ge 2),(6x-1", " x lt 2):}`

Answer

To discuss the continuity of the function

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

we need to check its continuity at the point \(x = 2\). A function is continuous at a point if the following three conditions are satisfied:

1. \(f(a)\) is defined.
2. \(\lim_{x \to a} f(x)\) exists.
3. \(\lim_{x \to a} f(x) = f(a)\).

Here, \(a = 2\).

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

For \(x = 2\), we use the first case of the function since \(x \geq 2\):

\[ f(2) = 3(2) + 5 = 6 + 5 = 11. \]

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

Next, we find the left-hand limit as \(x\) approaches 2 from the left (\(x < 2\)). We use the second case of the function:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (6x - 1). \]

Substituting \(x = 2\):

\[ \lim_{x \to 2^-} f(x) = 6(2) - 1 = 12 - 1 = 11. \]

### Step 3: Calculate \(\lim_{x \to 2^+} f(x)\)

Now, we find the right-hand limit as \(x\) approaches 2 from the right (\(x \geq 2\)). We use the first case of the function:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + 5). \]

Substituting \(x = 2\):

\[ \lim_{x \to 2^+} f(x) = 3(2) + 5 = 6 + 5 = 11. \]

### Step 4: Compare the values

Now we compare the three values we have calculated:

- \(f(2) = 11\)
- \(\lim_{x \to 2^-} f(x) = 11\)
- \(\lim_{x \to 2^+} f(x) = 11\)

Since all three values are equal, we conclude that:

\[ \lim_{x \to 2} f(x) = f(2). \]

### Conclusion

Thus, the function \(f(x)\) is continuous at \(x = 2\).

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