Question

at what points is the following function continuous?
$f(x) = \

$$\begin{cases} \\dfrac{x^2 - x - 2}{x - 2}, & x \ eq 2 \\\\ 3, & x = 2 \\end{cases}$$

$
the function is continuous on $\square$. (type your answer in interval notation)

Explanation:

Step1: Simplify the rational function for \( x

eq 2 \)
Factor the numerator \( x^2 - x - 2 \). We know that \( x^2 - x - 2=(x - 2)(x+ 1) \). So for \( x
eq2 \), the function \( \frac{x^2 - x - 2}{x - 2}=\frac{(x - 2)(x + 1)}{x - 2}=x + 1 \) (we can cancel \( x - 2 \) since \( x
eq2 \)).

Step2: Check continuity at \( x = 2 \)

First, find the limit as \( x \) approaches \( 2 \) of \( f(x) \). Since for \( x
eq2 \), \( f(x)=x + 1 \), then \( \lim_{x
ightarrow2}f(x)=\lim_{x
ightarrow2}(x + 1)=2+1 = 3 \). Now, check the value of the function at \( x = 2 \), which is \( f(2)=3 \). Since \( \lim_{x
ightarrow2}f(x)=f(2) \), the function is continuous at \( x = 2 \).

Step3: Check continuity for all real numbers

The function \( f(x) \) is a polynomial (after simplification for \( x
eq2 \), and continuous at \( x = 2 \)) and polynomials are continuous everywhere on \( (-\infty,\infty) \). So the function is continuous for all real numbers.

Answer:

\((-\infty, \infty)\)