Question

assignment 3: problem 11 (1 point) find the value of the constant c that makes the following function continuous on (-∞,∞). f(a) = {ca + 8 if a ∈ (-∞,6; ca² - 8 if a ∈ (6,∞)

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $a = 6$, $\lim_{a
ightarrow6^{-}}f(a)=\lim_{a
ightarrow6^{+}}f(a)$.

Step2: Calculate left - hand limit

$\lim_{a
ightarrow6^{-}}f(a)=c\times6 + 8=6c + 8$.

Step3: Calculate right - hand limit

$\lim_{a
ightarrow6^{+}}f(a)=c\times6^{2}-8 = 36c-8$.

Step4: Set left - hand and right - hand limits equal

$6c + 8=36c-8$.

Step5: Solve for $c$

First, move terms involving $c$ to one side: $36c-6c=8 + 8$. Then $30c=16$, so $c=\frac{16}{30}=\frac{8}{15}$.

Answer:

$\frac{8}{15}$