Show that the function
`f (x)={ {:(3x^(2) + 12 x - 1,- 1 le x le 2 ),(" "37 - x," "2 lt x le 3 ):}` is continuous at x = 2

Answer

To show that the function

\[ f(x) = \begin{cases} 3x^2 + 12x - 1 & \text{for } -1 \leq x \leq 2 \\ 37 - x & \text{for } 2 < x \leq 3 \end{cases} \]

is continuous at \( x = 2 \), we need to verify that:

1. \( f(2) \) is defined.
2. The left-hand limit as \( x \) approaches 2 exists and is equal to \( f(2) \).
3. The right-hand limit as \( x \) approaches 2 exists and is equal to \( f(2) \).

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

Since \( 2 \) falls within the first case of the piecewise function:

\[ f(2) = 3(2^2) + 12(2) - 1 \]

Calculating this:

\[ = 3(4) + 24 - 1 \]
\[ = 12 + 24 - 1 \]
\[ = 35 \]

### Step 2: Calculate the left-hand limit as \( x \) approaches 2

The left-hand limit is calculated using the first part of the function:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x^2 + 12x - 1) \]

Substituting \( x = 2 \):

\[ = 3(2^2) + 12(2) - 1 \]
\[ = 12 + 24 - 1 \]
\[ = 35 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 2

The right-hand limit is calculated using the second part of the function:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (37 - x) \]

Substituting \( x = 2 \):

\[ = 37 - 2 \]
\[ = 35 \]

### Conclusion

Now we have:

- \( f(2) = 35 \)
- \( \lim_{x \to 2^-} f(x) = 35 \)
- \( \lim_{x \to 2^+} f(x) = 35 \)

Since all three values are equal, we conclude that the function \( f(x) \) is continuous at \( x = 2 \).

\[ \text{Hence, } f(x) \text{ is continuous at } x = 2. \]

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