MathAdmin: Created page with " Let :: $f(x) = \left\{ \begin{array}{lr} \frac{x^2-2x-3}{x-3} & \text{if }x \ne 3\\ 5 & \text{if }x = 3 \end{array} \right...."$

2017-04-10T16:20:35Z

Created page with " Let ::<math>f(x) = \left\{ \begin{array}{lr} \frac{x^2-2x-3}{x-3} & \text{if }x \ne 3\\ 5 & \text{if }x = 3 \end{array} \right...."

New page

Let

::<math>f(x) = \left\{
\begin{array}{lr}
\frac{x^2-2x-3}{x-3} & \text{if }x \ne 3\\
5 & \text{if }x = 3
\end{array}
\right.
</math>
 For what values of  <math style="vertical-align: 0px">x</math>  is  <math style="vertical-align: -4px">f</math>  continuous?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|<math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: 0px">x=a</math>  if
|-
|       <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Based on the description of  <math style="vertical-align: -5px">f(x),</math> 
|-
|we know  <math style="vertical-align: -5px">f(x)</math>  is continuous on
|-
|       <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</math> 
|-
|Now, we need to see if  <math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: 0px">x=3.</math> 
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have
|-
|       <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} \frac{x^2-2x-3}{x-3}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3^+} \frac{(x-3)(x+1)}{x-3}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3^+} x+1}\\
&&\\
& = & \displaystyle{4.}
\end{array}</math>
|-
|Similarly,
|-
|       <math>\lim_{x\rightarrow 3^-}f(x)=4.</math>
|-
|Since
|-
|       <math style="vertical-align: 0px">\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow 3^+}f(x)</math> 
|-
|we have
|-
|       <math>\lim_{x\rightarrow 3}f(x)=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|But, since
|-
|       <math>f(3)=5,</math>
|-
|we have
|-
|       <math>\lim_{x\rightarrow 3}f(x)\ne f(3).</math>
|-
|Therefore,  <math style="vertical-align: -5px">f(x)</math>  is not continuous at  <math style="vertical-align: 0px">x=3.</math> 
|-
| <math style="vertical-align: -5px">f(x)</math>  is continuous only on  <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</math> 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math style="vertical-align: -5px">f(x)</math>  is continuous on  <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</math> 
|}
[[009A_Sample_Final_2|'''Return to Sample Exam''']]

009A Sample Final 2, Problem 2 - Revision history

MathAdmin: Created page with " Let ::f(x) = \left\{ \begin{array}{lr} \frac{x^2-2x-3}{x-3} & \text{if }x \ne 3\\ 5 & \text{if }x = 3 \end{array} \right...."

MathAdmin: Created page with " Let :: $f(x) = \left\{ \begin{array}{lr} \frac{x^2-2x-3}{x-3} & \text{if }x \ne 3\\ 5 & \text{if }x = 3 \end{array} \right...."$