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The piecewise function provided is:

\[
f(x) = 
\begin{cases} 
\sqrt{kx}, & 0 \leq x \leq 3 \\
x + 1, & 3 < x \leq 10
\end{cases}
\]

To ensure continuity at \( x = 3 \), the two parts of the function must have the same value at that point. Therefore, we need to set:

\[
\sqrt{k \cdot 3} = 3 + 1
\]

Now, let's solve for \( k \). First, simplify both sides of the equation:

\[
\sqrt{3k} = 4
\]

Square both sides:

\[
3k = 16
\]

Now solve for \( k \):

\[
k = \frac{16}{3}
\]

So, for the function to be continuous at \( x = 3 \), \( k = \frac{16}{3} \).

Would you like me to go into more details, or do you have any further questions?

Here are some related questions:
1. How do we check if a function is differentiable at a point where two pieces meet?
2. Can piecewise functions always be made continuous by adjusting parameters?
3. What does it mean for a function to be continuous at a point?
4. Can you find the derivative of this piecewise function?
5. How does the square root affect the domain of the function?

**Tip:** When solving for continuity in piecewise functions, always equate the boundary values of the two expressions at the connecting point!

Given the piecewise function: f(x) = { sqrt(kx), 0 ≤ x ≤ 3; x + 1, 3 < x ≤ 10 }, find the value of k that ensures continuity at x = 3.

Learn how to solve for the value of k in the piecewise function f(x) = sqrt(kx) for 0 ≤ x ≤ 3 and x + 1 for 3 < x ≤ 10 to ensure continuity at x = 3. This problem covers key mathematical concepts in piecewise functions and continuity, suitable for Grades 9-12.

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