Math 22 Continuity

Continuity

Informally, a function is continuous at $x=c$ means that there is no interruption in the graph of $f$ at $c$ .

Definition of Continuity

Let $c$ be a real number in the interval $(a,b)$ , and let $f$ be a function whose domain contains the interval $(a,b)$ . The function $f$ is continuous at $c$ when
these conditions are true.
1. $f(c)$ is defined.
2. $\lim _{x\to c}f(x)$ exists.
3. $\lim _{x\to c}f(x)=f(c)$ If $f$ is continuous at every point in the interval $(a,b)$ , then $f$ is continuous on the open interval $(a,b)$ .

Continuity of piece-wise functions

Discuss the continuity of $f(x)={\begin{cases}x+2&{\text{if }}-1\leq x<3\\14-x^{2}&{\text{if }}3\leq x\leq 5\end{cases}}$ Solution: On the interval $[-1,3)$ , $f(x)=x+2$ and it is a polynomial function so it is continuous on $[-1,3)$ On the interval $[3,5]$ , $f(x)=14-x^{2}$ and it is a polynomial function so it is continuous on $[3,5]$ Finally we need to check if $f(x)$ is continuous at $x=3$ . So, consider $\lim _{x\to 3^{-}}f(x)=\lim _{x\to 3^{-}}x+2=3+2=5$ Then, $\lim _{x\to 3^{+}}f(x)=\lim _{x\to 3^{+}}14-x^{2}=14-(3)^{2}=5$ . Since $\lim _{x\to 3^{-}}f(x)=5=\lim _{x\to 3^{+}}f(x)$ , \lim_{x\to 3} f(x) exists. Also notice $f(3)=14-(3)^{2}=5=\lim _{x\to 3}f(x)$ So by definition of continuity, $f(x)$ is continuous at $x=3$ . Hence, $f(x)$ is continuous on $[-1,5]$ 