Question

find the values of x for which the function $y = \frac{x + 3}{x^2 - 3x - 4}$ is discontinuous. a. $x = 3$ b. $x = -4, x = 1$ c. $x = -1, x = 4$ d. $x = -3$

Explanation:

Step1: Find denominator roots

Factor denominator: $x^2 - 3x - 4 = (x + 1)(x - 4)$
Set to zero: $(x + 1)(x - 4) = 0$

Step2: Solve for x

$x + 1 = 0 \Rightarrow x = -1$; $x - 4 = 0 \Rightarrow x = 4$
These x-values make the function undefined (discontinuous).

Answer:

C. x=-1, x=4