Determining Whether a Piecewise Function is Continuous at a Given Number

To determine if the function $f$ is continuous at $x=a,$ we will determine if the three conditions of continuity are satisfied at $x=a$.

• ⓐ

  Condition 1: Does $f(a)$ exist?

  $$\begin{array}{l}f(3)=4(3)=12\hfill \\ \Rightarrow \text{Condition 1 is satisfied}.\hfill \end{array}$$

  Condition 2: Does $\underset{x\to 3}{\lim }f(x)$ exist?

  To the left of $x=3,$ $f(x)=4x;$ to the right of $x=3,$ $f(x)=8+x.$ We need to evaluate the left- and right-hand limits as $x$ approaches 1.

  • Left-hand limit: $\underset{x\to 3^{-}}{\lim }f(x)=\underset{x\to 3^{-}}{\lim }4(3)=12$
  • Right-hand limit: $\underset{x\to 3^{+}}{\lim }f(x)=\underset{x\to 3^{+}}{\lim }\left(8+x\right)=8+3=11$

  Because $\underset{x\to 3^{-}}{\lim }f(x)\ne \underset{x\to 3^{+}}{\lim }f(x),$ $\underset{x\to 3}{\lim }f(x)$ does not exist.

  $$\Rightarrow \text{ Condition 2 fails}\text{.}$$

  There is no need to proceed further. Condition 2 fails at $x=3.$ If any of the conditions of continuity are not satisfied at $x=3,$ the function $f\left(x\right)$ is not continuous at $x=3.$

• ⓑ

  $x=\frac{8}{3}$

  Condition 1: Does $f\left(\frac{8}{3}\right)$ exist?

  $$\begin{array}{l}f\left(\frac{8}{3}\right)=4\left(\frac{8}{3}\right)=\frac{32}{3}\hfill \\ \Rightarrow \text{Condition 1 is satisfied}.\hfill \end{array}$$

  Condition 2: Does $\underset{x\to \frac{8}{3}}{\lim }f(x)$ exist?

  To the left of $x=\frac{8}{3},$ $f(x)=4x;$ to the right of $x=\frac{8}{3},$ $f(x)=8+x.$ We need to evaluate the left- and right-hand limits as $x$ approaches $\frac{8}{3}.$

  • Left-hand limit: $\underset{x\to {\frac{8}{3}}^{-}}{\lim }f(x)=\underset{x\to {\frac{8}{3}}^{-}}{\lim }4\left(\frac{8}{3}\right)=\frac{32}{3}$
  • Right-hand limit: $\underset{x\to {\frac{8}{3}}^{+}}{\lim }f(x)=\underset{x\to {\frac{8}{3}}^{+}}{\lim }\left(8+x\right)=8+\frac{8}{3}=\frac{32}{3}$

  Because $\underset{x\to \frac{8}{3}}{\lim }f(x)$ exists,

  $$\Rightarrow \text{Condition 2 is satisfied}.$$

  Condition 3: Is $f\left(\frac{8}{3}\right)=\underset{x\to \frac{8}{3}}{\lim }f(x)?$

  $$\begin{array}{l}f\left(\frac{32}{3}\right)=\frac{32}{3}=\underset{x\to \frac{8}{3}}{\lim }f(x)\hfill \\ \Rightarrow \text{Condition 3 is satisfied}.\hfill \end{array}$$

  Because all three conditions of continuity are satisfied at $x=\frac{8}{3},$ the function $f\left(x\right)$ is continuous at $x=\frac{8}{3}.$