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The problem describes a piecewise function \( f(x) \) defined on different intervals of \( x \):

1. \( f(x) = \frac{\sin(24x)}{x} \) for \( x < 0 \)
2. \( f(x) = x^2 + ax + b \) for \( 0 \leq x \leq 6 \)
3. \( f(x) = 18 \) for \( x > 6 \)

Given that the function is continuous, we need to ensure continuity at the boundaries of these intervals. The function's continuity at \( x = 0 \) implies:
\[ \lim_{x \to 0^-} f(x) = f(0) \]
\[ \lim_{x \to 0^-} \frac{\sin(24x)}{x} = f(0) \]
Using the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ \lim_{x \to 0^-} \frac{\sin(24x)}{x} = 24 \]

Thus, \( f(0) = 24 \), and for the quadratic part:
\[ 24 = 0^2 + a \cdot 0 + b \]
\[ b = 24 \]

Continuity at \( x = 6 \) requires:
\[ \lim_{x \to 6^+} f(x) = f(6) \]
\[ 18 = 6^2 + 6a + 24 \]
\[ 18 = 36 + 6a + 24 \]
\[ 18 = 60 + 6a \]
\[ 6a = -42 \]
\[ a = -7 \]

So, the complete function definition is:
\[ f(x) = \begin{cases} 
\frac{\sin(24x)}{x} & \text{for } x < 0 \\
x^2 - 7x + 24 & \text{for } 0 \leq x \leq 6 \\
18 & \text{for } x > 6 
\end{cases} \]

To find \( f(2) \):
\[ f(2) = 2^2 - 7 \times 2 + 24 \]
\[ f(2) = 4 - 14 + 24 \]
\[ f(2) = 14 \]

Therefore, \( f(2) = 14 \), and the correct answer is option (d) 14.

Do you have any other questions or need further explanation on any of the steps?

Here are five related questions you might find interesting:
1. What would the function value be at \( x = -1 \)?
2. How would the function definition change if \( f \) needed to be continuous at \( x = 7 \) instead?
3. Calculate \( f(6) \) directly from the quadratic equation.
4. What is the derivative of \( f(x) \) at \( x = 4 \)?
5. How would the values of \( a \) and \( b \) change if the function value at \( x = 6 \) were 20 instead of 18?

**Tip:** When checking for continuity at boundaries of piecewise functions, ensure that the limits from both sides of the boundary and the function value at the boundary itself are equal. This helps in finding any unknown constants in the function's expression.

Sejam a e b números reais e considere a função f : R -> R definida como: f(x) = sen(24x)/x, se x < 0; f(x) = x² + ax + b, se 0 <= x <= 6; f(x) = 18, se x > 6. Se a função f é contínua, então f(2) é igual a: (a) 2; (b) 20; (c) 18; (d) 14; (e) 6.

This problem explores the continuity of a piecewise function involving trigonometric limits and a quadratic equation. The solution guides through calculating f(2) by ensuring the function remains continuous at key intervals.

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