Question

Determine whether $f$is continuous at each of the following numbers. If it is not, explain why not.

$\left(a\right)x=-2\left(b\right)x=1\left(c\right)x=3\left(d\right)x=4$

Short answer

The function is continuous at $x=-2,4$and discontinuous at$x=1,3$

Step-by-step solution

Part (a) Step 1. Given Information

A graph is given.

We need to find if the function is continuous at$x=-2$

Part(a) Step 2. Explanation

For $x=2,f\left(2\right)=2$It is clear from the graph that,

$$\begin{gathered} \underset{x\to -2^{-}}{\lim }f\left(x\right)=2 \\ \underset{x\to -2^{+}}{lim}f\left(x\right)=2 \\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=2=\underset{x\to -2^{+}}{\lim }f\left(x\right) \\ \underset{x\to -2}{\lim }f\left(x\right)=2 \end{gathered}$$

Therefore, continuous limit exists.

f(1)=0Part(b) Step 1. Explanation

For $x=1$

It is clear from the graph that,

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=2 \\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=2 \\ \underset{x\to 1^{-}}{\lim }f\left(x\right)=2=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \underset{x\to 1}{\lim }f\left(x\right)=2Butf\left(1\right)=0 \end{gathered}$$

Therefore, continuous limit does not exist.

Part(c) Step 1. Explanation

For $x=3,f\left(3\right)=-3$

It is clear from the graph,

$$\begin{gathered} \underset{x\to 3^{-}}{\lim }f\left(x\right)=5 \\ \underset{x\to 3^{+}}{\lim }f\left(x\right)=-3 \end{gathered}$$

Since,

$\underset{x\to 3^{-}}{\lim }f\left(x\right)=5\ne -3=\underset{x\to 3^{+}}{\lim }f\left(x\right)$

Therefore, continuous limit does not exist.

Part(d) Step 1. Explanation

For $x=4,f\left(4\right)=-4$

From the graph,

$\underset{x\to 4}{\lim }f\left(x\right)=-4$

Therefore, continuous limit exists.