Question

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

$$\begin{cases}\\frac{x^{2}-2x - 3}{x - 3},&x\ eq3\\4,&x = 3\\end{cases}$$

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

Explanation:

Step1: Simplify the function for $x

eq3$
Factor the numerator: $\frac{x^{2}-2x - 3}{x - 3}=\frac{(x - 3)(x+1)}{x - 3}=x + 1$ for $x
eq3$.

Step2: Check the limit as $x$ approaches 3

$\lim_{x
ightarrow3}\frac{x^{2}-2x - 3}{x - 3}=\lim_{x
ightarrow3}(x + 1)=3+1 = 4$.

Step3: Check the value of the function at $x = 3$

$f(3)=4$. Since $\lim_{x
ightarrow3}f(x)=f(3)=4$, the function is continuous at $x = 3$. And the function $y=x + 1$ is a linear - function which is continuous for all real numbers.

Answer:

$(-\infty,\infty)$