Q. 10 - Calculus

Question

Each function in Exercises 9–12 is discontinuous at some value x = c. Describe the type of discontinuity and any one-sided continuity at x = c, and sketch a possible graph of f. 

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=2, \underset{x\to 2^{+}}{\lim }f\left(x\right)=1, f\left(2\right)=1.$

Step-by-Step Solution

Answer

The type of discontinuity is a jump discontinuity and f(x) is right continuous at $x=2.$

The graph of f is 

Step 1. Given Information.

The given function is $\underset{x\to 2^{-}}{\lim }f\left(x\right)=2, \underset{x\to 2^{+}}{\lim }f\left(x\right)=1, f\left(2\right)=1.$

Step 2. Describing the discontinuity.

From the function, we can depict that $\underset{x\to 2^{-}}{\lim }f\left(x\right)=2 \mathrm{and} \underset{\mathrm{x}\to 2^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1$ both exist but are not equal to $f\left(2\right)=1.$

Thus, f(x) has a jump discontinuity at $x=2.$

Step 3. Describing one-sided continuity at x=c .

The $f\left(x\right)$ is right continuous at $x=2$ but not left continuous because $\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right).$

Step 4. Graph of f .

The graph of f is