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 $f(x)={\begin{cases}x+2&{\text{if }}-1\leq x<3\\14-x^{2}&{\text{if }}3\leq x\leq 5\end{cases}}$

ExpandSolution:
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 $[-1,5]$

Notes

Polynomial function is continuous on the entire real number line (ex: $f(x)=2x^{2}-1$ is continuous on $(-\infty ,\infty )$)

Rational functions is continuous at every number in its domain. (ex: $f(x)={\frac {x+2}{x^{2}-1}}$ is continuous on $(-\infty ,-1)\cup (-1,1)\cup (1,\infty )$ since the denominator cannot equal to zero)

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