Find the value of k for which the function ` f(x)={{:(3x-8, if x le 5),(2k, if x gt 5) :}` is continuous at `x = 5`

Explanation

To find the value of \( k \) for which the function

\[ f(x) = \begin{cases} 3x - 8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases} \]

is continuous at \( x = 5 \), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at that point are equal.

### Step 1: Find the left-hand limit as \( x \) approaches 5
The left-hand limit is given by:

\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (3x - 8) \]

Substituting \( x = 5 \):

\[ = 3(5) - 8 = 15 - 8 = 7 \]

### Step 2: Find the right-hand limit as \( x \) approaches 5
The right-hand limit is given by:

\[ \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (2k) = 2k \]

### Step 3: Find the value of the function at \( x = 5 \)
Since \( x = 5 \) falls under the condition \( x \leq 5 \), we have:

\[ f(5) = 3(5) - 8 = 15 - 8 = 7 \]

### Step 4: Set the limits equal to each other
For the function to be continuous at \( x = 5 \), we need:

\[ \text{Left-hand limit} = \text{Right-hand limit} = f(5) \]

This gives us:

\[ 7 = 2k = 7 \]

### Step 5: Solve for \( k \)
From the equation \( 2k = 7 \):

\[ k = \frac{7}{2} \]

### Conclusion
The value of \( k \) for which the function is continuous at \( x = 5 \) is:

\[ \boxed{\frac{7}{2}} \]