Question

Omain. In the practice exam question here, x = 2 is the chosen value, of course!

f(x) =
[(2x + 1)(x-2)
x=2
k
for x = 2
for x = 2
9. Let f be the function defined above. For what value of k is f continuous at x = 2?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 5
ANSWER

Answer 1

To determine the value of k that would make the function f continuous at x = 2, we need to evaluate the function at x = 2 and ensure that the left-hand limit and the right-hand limit are equal at that point.

Given the function:
f(x) = [(2x + 1)(x - 2)] / (x - 2)

Let's substitute x = 2 into the function:
f(2) = [(2(2) + 1)(2 - 2)] / (2 - 2)
= [(4 + 1)(0)] / 0

At this point, we have a division by zero, which indicates that the function is undefined at x = 2. In order for the function to be continuous at x = 2, we need to find a value of k that eliminates the discontinuity.

Looking at the expression, we can see that (x - 2) appears in both the numerator and denominator. To eliminate the discontinuity at x = 2, we can cancel out the (x - 2) term.

Therefore, k should be the value that makes (x - 2) cancel out. In this case, k = 5. When k = 5, the (x - 2) terms cancel out, resulting in a simplified expression of f(x) = 2x + 1.

Hence, the correct answer is (E) 5.