1. **Problem Statement:** Find the value of $k$ such that the function $$f(x) = \begin{cases} x^2 - 1 & \text{for } x < 3 \\ 2kx & \text{for } x \geq 3 \end{cases}$$ is continuous at $x = 3$. 2. **Continuity Condition:** A function $f(x)$ is continuous at $x = a$ if $$\lim_{x \to a^-} f(x) = f(a) = \lim_{x \to a^+} f(x).$$ 3. **Apply the condition at $x=3$: ** - Left-hand limit: $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 - 1) = 3^2 - 1 = 9 - 1 = 8.$$ - Right-hand limit and value at $x=3$: $$f(3) = 2k \times 3 = 6k.$$ 4. **Set the limits equal for continuity:** $$8 = 6k.$$ 5. **Solve for $k$: ** $$k = \frac{8}{6} = \frac{4}{3}.$$ **Final answer:** $$\boxed{k = \frac{4}{3}}.$$