at what points is the following function continuous?\nf(x) = \\begin{cases}\\frac{x^{2}-2x - 3}{x…

at what points is the following function continuous?\nf(x) = \\begin{cases}\\frac{x^{2}-2x - 3}{x - 3},&x\\neq3\\4,&x = 3\\end{cases}\nthe function is continuous on . (type your answer in interval notation.)

at what points is the following function continuous?\nf(x) = \\begin{cases}\\frac{x^{2}-2x - 3}{x - 3},&x\\neq3\\4,&x = 3\\end{cases}\nthe function is continuous on . (type your answer in interval notation.)

Answer

Explanation:

Step1: Simplify the function for $x\neq3$

Factor the numerator: $\frac{x^{2}-2x - 3}{x - 3}=\frac{(x - 3)(x+1)}{x - 3}=x + 1$ for $x\neq3$.

Step2: Check the limit as $x$ approaches 3

$\lim_{x\rightarrow3}\frac{x^{2}-2x - 3}{x - 3}=\lim_{x\rightarrow3}(x + 1)=3+1 = 4$.

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

$f(3)=4$. Since $\lim_{x\rightarrow3}f(x)=f(3)=4$, 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:

$(-\infty,\infty)$