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We are given a piecewise function and need to determine the value of \( k \) that makes the function continuous. The function is defined as follows:

\[
f(x) = 
\begin{cases} 
kx, & \text{if } 0 \leq x < 2 \\
2k, & \text{if } 2 \leq x < 4 \\
-kx + 5, & \text{if } x \geq 4 
\end{cases}
\]

To ensure the function is continuous, the left-hand limit must equal the right-hand limit at the points where the function changes, i.e., at \( x = 2 \) and \( x = 4 \).

### Step 1: Continuity at \( x = 2 \)
We need to check if the values of \( f(x) \) as \( x \) approaches 2 from both sides are equal:

1. From the left (\( 0 \leq x < 2 \)):
   \[
   \lim_{x \to 2^-} f(x) = k(2) = 2k
   \]

2. From the right (\( 2 \leq x < 4 \)):
   \[
   \lim_{x \to 2^+} f(x) = 2k
   \]

Since both limits are already equal at \( x = 2 \), there is no restriction here.

### Step 2: Continuity at \( x = 4 \)
Now, we check if the values of \( f(x) \) as \( x \) approaches 4 from both sides are equal:

1. From the left (\( 2 \leq x < 4 \)):
   \[
   \lim_{x \to 4^-} f(x) = 2k
   \]

2. From the right (\( x \geq 4 \)):
   \[
   \lim_{x \to 4^+} f(x) = -4k + 5
   \]

For the function to be continuous at \( x = 4 \), these two expressions must be equal:
\[
2k = -4k + 5
\]

### Step 3: Solve for \( k \)
Now, solve the equation:
\[
2k + 4k = 5
\]
\[
6k = 5
\]
\[
k = \frac{5}{6}
\]

Thus, the value of \( k \) that makes the function continuous is \( \frac{5}{6} \).

The correct answer should be checked in the given choices. However, none of the given choices (1, 2, 3, 5) are correct, since \( k = \frac{5}{6} \). There seems to be an issue with the provided answer options.

Would you like more details or have any other questions? Here are some related questions to expand on this topic:

1. What is the general condition for continuity of a piecewise function?
2. How do you check for continuity at different points in a piecewise function?
3. Can a piecewise function be continuous if \( k \) values are not equal at all points?
4. What happens if the left-hand limit and right-hand limit are not equal at a given point?
5. How does the behavior of a piecewise function affect its differentiability?

**Tip:** Always check the function values at boundary points to ensure both the left and right limits match, ensuring continuity.

For the piecewise function f(x) = {kx, if 0 ≤ x < 2; 2k, if 2 ≤ x < 4; -kx + 5, if x ≥ 4}, determine which value of k makes f(x) continuous.

Learn how to determine the value of k that makes a piecewise function continuous by solving for the limits at boundary points x = 2 and x = 4. This detailed explanation walks through step-by-step calculations for piecewise functions, focusing on continuity and limit matching.

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