Determine whether the function is continuous at the given x-value. If discontinuous, state the type of discontinuity (jump, inf., removable)?

Question

1. y=(3)/(2x^2)

at x=-1

2. y=(x-2)/(x-4)

at x=4

3. y=(x+2)/(x^2+6x+8)

at x=-2

4. y={x^2+1, if x<-3     -x-3, if x≥-3}

at x=2

5. y={x^2+1, if x<-3    -x-3, if x≥-3}

at x=-3

Answer 1

1.) Continuous, the function is defined at x = -1.

2.) Infinite discontinuity, There is a vertical asymptote at x = 4.

3.) Removable discontinuity, Factor the denominator into (x + 4)(x + 2), and (x+2) cancels out with the numerator, leaving no discontinuity.

4.) Continuous, we need to check only the right half of the domain, and -2 - 3 = -5.

5.) Jump discontinuity, we plug into both pieces and get different values (10 and 0).