find the value of the constant k that makes the function continuous.\ng(x) = {\\(\\frac{2x^{2}-9x - 18}{x…

find the value of the constant k that makes the function continuous.\ng(x) = {\\(\\frac{2x^{2}-9x - 18}{x - 6}\\) if x≠6, kx - 9 if x = 6}\nwrite an equation that can be solved to find k.\na. \\(\\lim_{x\\to6}\\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9\\)\nb. \\(\\frac{2(6)^{2}-9(6)-18}{6 - 6}=\\lim_{x\\to6}k(x)-9\\)\nc. \\(\\lim_{x\\to6}\\frac{2x^{2}-9x - 18}{x - 6}=k\\)\nd. \\(\\frac{2(6)^{2}-9(6)-18}{6 - 6}=k(6)-9\\)

find the value of the constant k that makes the function continuous.\ng(x) = {\\(\\frac{2x^{2}-9x - 18}{x - 6}\\) if x≠6, kx - 9 if x = 6}\nwrite an equation that can be solved to find k.\na. \\(\\lim_{x\\to6}\\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9\\)\nb. \\(\\frac{2(6)^{2}-9(6)-18}{6 - 6}=\\lim_{x\\to6}k(x)-9\\)\nc. \\(\\lim_{x\\to6}\\frac{2x^{2}-9x - 18}{x - 6}=k\\)\nd. \\(\\frac{2(6)^{2}-9(6)-18}{6 - 6}=k(6)-9\\)

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = a$, $\lim_{x\rightarrow a}g(x)=g(a)$. Here $a = 6$, so $\lim_{x\rightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9$.

Answer:

A. $\lim_{x\rightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9$