Question

find the value of the constant k that makes the function continuous.
g(x) = {\\(\frac{2x^{2}-9x - 18}{x - 6}\\) if x≠6, kx - 9 if x = 6}
write an equation that can be solved to find k.
a. \\(\lim_{x\to6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9\\)
b. \\(\frac{2(6)^{2}-9(6)-18}{6 - 6}=\lim_{x\to6}k(x)-9\\)
c. \\(\lim_{x\to6}\frac{2x^{2}-9x - 18}{x - 6}=k\\)
d. \\(\frac{2(6)^{2}-9(6)-18}{6 - 6}=k(6)-9\\)

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a}g(x)=g(a)$. Here $a = 6$, so $\lim_{x
ightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9$.

Answer:

A. $\lim_{x
ightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9$