Question

Find the x-values (if any) at which f is not continuous. Which of the discontinuities are removable (holes) and which are non-removable (asymptotes)? If the function is continuous, write NONE for holes and asymptotes.

f(x) = 2x² - 6x
x² - x - 6
Continuous or Not continuous at:
Holes:
Asymptotes:

Answer 1

The function is not continuous at x = 3 and x = -2. At x = 3, there is a removable discontinuity (hole), and at x = -2, there is a non-removable discontinuity (asymptote).

Step-by-step explanation:

To find the x-values at which f is not continuous, we need to identify any points where the function is undefined. In this case, the function f(x) = 2x² - 6x / x² - x - 6. To determine where the function is undefined, we set the denominator equal to zero and solve: x² - x - 6 = 0. Factoring the quadratic equation, we have (x - 3)(x + 2) = 0. This equation has two solutions: x = 3 and x = -2.

So, the function is not continuous at x = 3 and x = -2, as these values make the denominator zero. At x = 3, the function has a removable discontinuity, also known as a hole, because the numerator and denominator have a common factor of (x - 3). At x = -2, the function has a non-removable discontinuity, also known as an asymptote, because the numerator and denominator do not have a common factor that can be canceled out.