Chapter 2.4, Problem 2QQ

To determine

Tochoose:The false statement for the givenfunction $f\left(x\right)=\{\begin{array}{l}\frac{3}{4}x,0\le x<4\\ 2,x=4\\ -x+7,4<x\le 6\\ 1,6<x<8\end{array}$ .

Answer to Problem 2QQ

The correct choice is (E).

Explanation of Solution

Given information:The function is $f\left(x\right)=\{\begin{array}{l}\frac{3}{4}x,0\le x<4\\ 2,x=4\\ -x+7,4<x\le 6\\ 1,6<x<8\end{array}$ .

The left hand limit at $x=4$ is:

  $\begin{array}{c}\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{-}}{\lim }\left(\frac{3}{4}x\right)\\ =\frac{3}{4}\times 4\\ =3\end{array}$

The right hand limit at $x=4$ is:

  $\begin{array}{c}\underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }\left(-x+7\right)\\ =-4+7\\ =3\end{array}$

The value of $\underset{x\to 4^{-}}{\lim }f\left(x\right)$ and $\underset{x\to 4^{+}}{\lim }f\left(x\right)$ both are equal to $3$ .So, the value of $\underset{x\to 4}{\lim }f\left(x\right)$ exists and is equal to $3$ .

The value of $f\left(x\right)$ at $x=4$ is equal to $2$ .

  $f\left(4\right)=2$

The value of $\underset{x\to 4}{\lim }f\left(x\right)$ is not equal to value $f\left(4\right)$ .So, the function $f\left(x\right)$ is not continuous at $x=4$ .

Therefore, the correct choice is (E).