Question

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

• A. \(\frac27\)
• B. \(\frac72\)
• C. \(\frac37\)
• D. \(\frac47\)

Answer 1

The Correct Option is B

To determine the value of \(k\) for the function \( f(x) = \begin{cases} 3x - 8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases} \) to be continuous at \(x = 5\), we must ensure that the left-hand limit, the right-hand limit, and the function value at \(x = 5\) are all equal.

Step 1: Calculate the left-hand limit as \(x \to 5^-\)
For \(x \leq 5\), \(f(x) = 3x - 8\). So: \[ \lim_{x \to 5^-} f(x) = 3 \cdot 5 - 8 = 15 - 8 = 7. \]

Step 2: Calculate the right-hand limit as \(x \to 5^+\)
For \(x > 5\), \(f(x) = 2k\). Hence: \[ \lim_{x \to 5^+} f(x) = 2k. \]

Step 3: Enforce continuity at \(x = 5\)
We require: \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5). \] This gives: \[ 7 = 2k. \]

Step 4: Solve for \(k\)
\[ k = \frac{7}{2}. \]

Final Answer: \(\boxed{\frac{7}{2}}\)