12.3 Continuity  (Page 4/10)

To determine if the function $f$ is continuous at $x=5,$ we will determine if the three conditions of continuity are satisfied at $x=5.$

Condition 1:

There is no need to proceed further. Condition 2 fails at $x=5.$ If any of the conditions of continuity are not satisfied at $x=5,$ the function $f$ is not continuous at $x=5.$

The piecewise function is defined by three functions, which are all polynomial functions, $f(x)=x+1$ on $x<2,$ $f(x)=3$ on $2\le x<4,$ and $f(x)=x^2-5$ on $x\ge 4.$ Polynomial functions are continuous everywhere. Any discontinuities would be at the boundary points, $x=2$ and $x=4.$

At $x=2,$ let us check the three conditions of continuity.

Condition 1:

Condition 2: Because a different function defines the output left and right of $x=2,$ does $\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }f(x)?$

• Left-hand limit: $\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{-}}{\lim }\left(x+1\right)=2+1=3$
• Right-hand limit: $\underset{x\to 2^{+}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }3=3$

Because $3=3$ , $\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }f(x)$

Condition 3:

Because all three conditions are satisfied at $x=2,$ the function $f\left(x\right)$ is continuous at $x=2.$

At $x=4,$ let us check the three conditions of continuity.

Condition 2: Because a different function defines the output left and right of $x=4,$ does $\underset{x\to 4^{-}}{\lim }f(x)=\underset{x\to 4^{+}}{\lim }f(x)?$

• Left-hand limit: $\underset{x\to 4^{-}}{\lim }f(x)=\underset{x\to 4^{-}}{\lim }3=3$
• Right-hand limit: $\underset{x\to 4^{+}}{\lim }f(x)=\underset{x\to 4^{+}}{\lim }\left(x^2-11\right)=4^2-11=5$

Because $3\ne 5$ , $\underset{x\to 4^{-}}{\lim }f(x)\ne \underset{x\to 4^{+}}{\lim }f(x),$ so $\underset{x\to 4}{\lim }f(x)$ does not exist.

Because one of the three conditions does not hold at $x=4,$ the function $f(x)$ is discontinuous at $x=4.$