Question

Solve this very urgent​

Answer 1

Question: determine the value of k for which the function is continuous at x = 3.

f(x) = ((x + 3)² - 36)/(x - 3) , x ≠ 3

= k , x = 3

Answer:

12

Step-by-step explanation:

(x + 3)² - 36 = (x + 3)² - 6²

(x + 3)² - 36 = (x + 3 + 6)(x + 3 - 6)

(x + 3)² - 36 = (x + 9)(x - 3)

Therefore, for x ≠ 3

f(x) = ((x + 3)² - 36)/(x - 3)

f(x) = (x + 9)(x - 3)/(x - 3)

f(x) = x + 9

Notice that, for f(x) to be continuous at x = 3,

Lt (x tends to 3) f(x) = f(3)

=> Lt (x tends to 3) x + 9 = k

=> 3 + 9 = k

=> 12 = k