Chapter 14, Problem 36RE

To determine

The functionis continuous or not at the value of $x$

Answer to Problem 36RE

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

  $\underset{x\to 6^{-}}{\lim }f\left(x\right)=\text{continuous},\underset{x\to 6^{+}}{\lim }f\left(x\right)\text{=continuous},f\left(6\right)=5$

Explanation of Solution

Given:

The function $\underset{x\to -2^{-}}{\lim }f\left(x\right)=\text{continuous and}\underset{x\to 6^{-}}{\lim }f\left(x\right)=\text{continuous},f\left(-2\right)\text{and}f\left(6\right)$

Concept used:

The left continuity and right continuity of a function $f$ at a point $x=c$ are , equal

  $\begin{array}{l}\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)\\ \underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)\end{array}$

Calculation:

The function $\underset{x\to -2^{-}}{\lim }f\left(x\right)=\text{continuous},f\left(-2\right)=2$

Putting $x=-2$ in equation (1)

  $f\left(-2\right)=2$

The left continuity and right continuity of a function $f$ at a point $x=c$ are , equal

  $\begin{array}{l}\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)\\ \underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)\end{array}$

Left continuity

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

Right continuity

  $\underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$

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

  $\underset{x\to 6^{-}}{\lim }f\left(x\right)=\text{continuous},\underset{x\to 6^{+}}{\lim }f\left(x\right)\text{=continuous},f\left(6\right)=5$

This function is discontinuity of $f$ at $x=-2$

This is the jump discontinuity and any one sided discontinuity

So , function is not continuous at that given point