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\\) k = □

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = a\), \(\lim_{x
ightarrow a}g(x)=g(a)\). Here \(a = 6\), \(g(x)=\frac{2x^{2}-9x - 18}{x - 6}\) for \(x
eq6\) and \(g(x)=kx - 9\) for \(x = 6\). So \(\lim_{x
ightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9\).

Step2: Factor the numerator

Factor \(2x^{2}-9x - 18=(2x + 3)(x - 6)\). Then \(\lim_{x
ightarrow6}\frac{(2x + 3)(x - 6)}{x - 6}=\lim_{x
ightarrow6}(2x+3)\).

Step3: Evaluate the limit

\(\lim_{x
ightarrow6}(2x + 3)=2\times6+3=12 + 3=15\).

Step4: Solve for \(k\)

Set \(15=6k-9\). Add 9 to both sides: \(15 + 9=6k\), so \(24=6k\). Divide both sides by 6, we get \(k = 4\).

Answer:

A. \(\lim_{x
ightarrow6}\frac{2x^{2}-9x - 18}{x - 6}=k(6)-9\)
\(k = 4\)