Question

which function has a discontinuity when $x = -2$?

• $f(x) = \frac{x + 2}{x - 2}$
• $f(x) = \frac{x + 2}{x^2 + 4}$
• $f(x) = |x + 2| - 4$
• $f(x) = \

$$\begin{cases}2x - 1, & x < -2 \\\\ x^2 + 2, & x > -2\\end{cases}$$

$

Explanation:

Step1: Check discontinuity for Option 1

A rational function is discontinuous where denominator is 0. For $f(x)=\frac{x+2}{x-2}$, set $x-2=0 \implies x=2$. No discontinuity at $x=-2$.

Step2: Check discontinuity for Option 2

For $f(x)=\frac{x+2}{x^2+4}$, denominator $x^2+4$ is always positive ($x^2\geq0 \implies x^2+4\geq4$). No discontinuity at $x=-2$.

Step3: Check discontinuity for Option 3

The function $f(x)=|x+2|-4$ is an absolute value function, which is continuous everywhere, including $x=-2$.

Step4: Check discontinuity for Option 4

For piecewise function $f(x)=

$$\begin{cases}2x-1, & x < -2 \\x^2+2, & x > -2\end{cases}$$

$, there is no defined value at $x=-2$, and the function has a break here. This creates a discontinuity at $x=-2$.

Answer:

$f(x)=

$$\begin{cases}2x-1, & x < -2 \\x^2+2, & x > -2\end{cases}$$

$