let f be the function defined below. for what values of k is f continuous at x = 6? f(x)={\\frac{x^{2}-2x…

Explanation:

Step1: Factor the rational - function for (x\neq6)

For (x\neq6), (f(x)=\frac{x^{2}-2x - 24}{x^{2}-5x - 6}). Factor the numerator and denominator. The numerator (x^{2}-2x - 24=(x - 6)(x + 4)) and the denominator (x^{2}-5x - 6=(x - 6)(x+1)). So (f(x)=\frac{(x - 6)(x + 4)}{(x - 6)(x + 1)}=\frac{x + 4}{x + 1}) for (x\neq6).

Step2: Find the limit as (x\to6)

(\lim_{x\to6}\frac{x + 4}{x + 1}=\frac{6 + 4}{6+1}=\frac{10}{7}).

Step3: Use the definition of continuity

A function (y = f(x)) is continuous at (x=a) if (\lim_{x\to a}f(x)=f(a)). Since (f(x)) is continuous at (x = 6), and (\lim_{x\to6}f(x)=\frac{10}{7}), then (f(6)=k). So (k=\frac{10}{7}).

Answer:

(k=\frac{10}{7})