Chapter 11.3, Problem 90E

Solution

a)

To plot: All the possible graphs for $f$ .

All the possible graphs for $f\left(x\right)=\left\{\begin{array}{c}{x}^2-3x+3,\text{if}x\ne 2\\ a,\text{if}x=2\end{array}\right.$

Given information:

The given equation is $f\left(x\right)=\left\{\begin{array}{c}{x}^2-3x+3,\text{if}x\ne 2\\ a,\text{if}x=2\end{array}\right.$ .

Calculation:

Plot the graph for the function $f$ for $a=-1$ .

  

Plot the graph for the function $f$ for $a=1$ .

  

Plot the graph for the function $f$ for $a=3$ .

  

Therefore, several possible graphs for the function are plotted.

b)

To find: The value of $a$ resulting in continuous function $f$ at $x=2.1$ .

The value of $a$ is $1$ .

Given information:

The given equation is $f\left(x\right)=\left\{\begin{array}{c}{x}^2-3x+3,\text{if}x\ne 2\\ a,\text{if}x=2\end{array}\right.$ .

Formula used:

For a function to be continuous, $\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$ .

Calculation:

Determine the left limit for the function $f$ .

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }{x}^2-3x+3\\ =1\end{array}$

Determine the right limit for the function $f$ .

  $\begin{array}{c}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }{x}^2-3x+3\\ =1\end{array}$

Substitute $2$ for $x$ to find $f\left(2\right)$ .

  $\begin{array}{c}\underset{x\to 2}{\lim }f\left(x\right)=1\\ =f\left(2\right)\\ =a\end{array}$

Therefore, value of $a$ is $1$ resulting in continuous function $f$ at $x=2.1$ .