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The problem asks to determine the coefficient \( a \) so that the function \( f(x) \) is continuous at \( x = 0 \). The function is piecewise-defined as:

\[
f(x) = \begin{cases} 
x^2 + 6a, & x \leq 0 \\
\frac{\sin(6x)}{3x}, & x > 0
\end{cases}
\]

For the function to be continuous at \( x = 0 \), the left-hand limit (as \( x \) approaches 0 from the left) must equal the right-hand limit (as \( x \) approaches 0 from the right), and both must equal the function value at \( x = 0 \).

### Step 1: Evaluate the left-hand limit
For \( x \leq 0 \), the function is \( f(x) = x^2 + 6a \). So, the limit as \( x \to 0^- \) is:
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^2 + 6a) = 6a
\]

### Step 2: Evaluate the right-hand limit
For \( x > 0 \), the function is \( f(x) = \frac{\sin(6x)}{3x} \). To find the limit as \( x \to 0^+ \), we use the fact that \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \). Here, \( k = 6 \), so:
\[
\lim_{x \to 0^+} \frac{\sin(6x)}{3x} = \frac{6}{3} = 2
\]

### Step 3: Set the limits equal for continuity
For the function to be continuous at \( x = 0 \), the left-hand and right-hand limits must be equal:
\[
6a = 2
\]

### Step 4: Solve for \( a \)
\[
a = \frac{2}{6} = \frac{1}{3}
\]

Thus, the coefficient \( a \) must be \( \frac{1}{3} \) for the function to be continuous at \( x = 0 \).

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. What is the general definition of continuity at a point?
2. How would the process change if the piecewise function had a different form for \( x > 0 \)?
3. Can you provide an example of a function that is not continuous at a point?
4. What is the importance of limits in determining the continuity of a function?
5. How does the squeeze theorem apply to functions like \( \frac{\sin(kx)}{x} \) as \( x \to 0 \)?

**Tip:** When solving continuity problems with trigonometric functions, remember the standard limits involving sine and cosine, such as \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \).

Tentukan koefisien a agar fungsi f(x) kontinu di x = 0 dengan f(x) didefinisikan sebagai: f(x) = {x^2 + 6a, x ≤ 0; (sin(6x))/(3x), x > 0}.

Find the coefficient a so that the piecewise function f(x), defined by x^2 + 6a for x ≤ 0 and (sin(6x))/(3x) for x > 0, is continuous at x = 0. This solution covers key concepts in continuity, limits, and piecewise functions. Suitable for Grades 11-12.

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