Answer:
k = 1
Step-by-step explanation:
Given piecewise function:
[tex]f(x)=\begin{cases}x+8\quad x < 4\\3kx\quad \;\;\:x\geq 4\end{cases}[/tex]
Piecewise functions have multiple pieces of curves/lines where each piece corresponds to its definition over an interval.
To ensure the function is continuous at x = 4, the values of the function from the left side x < 4 and the right side x ≥ 4 must match at x = 4:
[tex]\displaystyle \lim_{{x \to 4^-}} f(x) = \lim_{{x \to 4^+}} f(x)[/tex]
For x < 4, the expression is x + 8, and for x ≥ 4, the expression is 3kx. Therefore:
[tex]\displaystyle \lim_{{x \to 4^-}} (x + 8) = \lim_{{x \to 4^+}} (3kx)[/tex]
Evaluate the left-hand limit:
[tex]\displaystyle \lim_{{x \to 4^-}} (x + 8) = 4 + 8 = 12[/tex]
Evaluate the right-hand limit:
[tex]\displaystyle \lim_{{x \to 4^+}} (3kx) = 3k \cdot 4 = 12k[/tex]
For continuity, these two limits must be equal:
[tex]12 = 12k[/tex]
Now, solve for k:
[tex]k = \dfrac{12}{12} = 1[/tex]
Therefore, the value of k that makes the function continuous at x = 4 is:
[tex]\Large\boxed{\boxed{k=1}}[/tex]
