009A Sample Final 1, Problem 2 - Revision history

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MathAdmin: Created page with " Consider the following piecewise defined function: :::::: $f(x) = \left\{ \begin{array}{lr} x+5 & \text{if }x < 3\\ 4\sqrt{x+1} & \..."$

2016-03-05T07:05:56Z

Created page with " Consider the following piecewise defined function: ::::::<math>f(x) = \left\{ \begin{array}{lr} x+5 & \text{if }x < 3\\ 4\sqrt{x+1} & \..."

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Consider the following piecewise defined function:

::::::<math>f(x) = \left\{
\begin{array}{lr}
x+5 & \text{if }x < 3\\
4\sqrt{x+1} & \text{if }x \geq 3
\end{array}
\right.
</math>
a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3</math>.

b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <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>
|-
|'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math>  is  <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math> We have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\
&&\\
& = & \displaystyle{4\sqrt{3+1}}\\
&&\\
& = & \displaystyle{8.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math> We have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\
&&\\
& = & \displaystyle{3+5}\\
&&\\
& = & \displaystyle{8.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we calculate <math style="vertical-align: -5px">f(3).</math> We have
|-
|
::<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
|-
|Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>  is continuous.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We need to use the limit definition of derivative and calculate the limit from both sides. So, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{h}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^-}1}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4\sqrt{3+h+1}-8}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})(\sqrt{4+h}+\sqrt{4})}{h(\sqrt{4+h}+\sqrt{4})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(4+h-4)}{h(\sqrt{4+h}+\sqrt{4})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4h}{h(\sqrt{4+h}+\sqrt{4})}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4}{(\sqrt{4+h}+\sqrt{4})}}\\
&&\\
& = & \displaystyle{\frac{4}{2\sqrt{4}}}\\
&&\\
& = & \displaystyle{1.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
|-
|<math style="vertical-align: -5px">f(x)</math>  is differentiable at <math style="vertical-align: 0px">x=3.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>  is continuous.
|-
|'''(b)''' Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable at <math style="vertical-align: 0px">x=3.</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the following piecewise defined function:
-::::::<math>f(x) = \left\{
+::<math>f(x) = \left\{
       \begin{array}{lr}
         x+5 &  \text{if }x < 3\\
@@ Line 8: / Line 8: @@
     \right.
 </math>
-::<span class="exam">a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3.</math>
+<span class="exam">(a) Show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3.</math>
-::<span class="exam">b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3.</math>
+<span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that &nbsp;<math style="vertical-align: -3px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.''' &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=a</math>&nbsp; if
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
-::'''1.''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous at <math style="vertical-align: 0px">x=a</math>&thinsp; if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
+|-
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
 |}
 '''Solution:'''
@@ Line 31: / Line 32: @@
 !Step 1: &nbsp;
 |-
-|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math> We have
+|We first calculate &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math>&nbsp; We have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\
 &&\\
@@ Line 46: / Line 47: @@
 !Step 2: &nbsp;
 |-
-|Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math> We have
+|Now, we calculate &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math>&nbsp; We have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\
 &&\\
@@ Line 61: / Line 62: @@
 !Step 3: &nbsp;
 |-
-|Now, we calculate <math style="vertical-align: -5px">f(3).</math> We have
+|Now, we calculate &nbsp;<math style="vertical-align: -5px">f(3).</math>&nbsp; We have
 |-
+|
-::<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>f(3)=4\sqrt{3+1}\,=\,8.</math>
 |-
-|Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
+|Since
 |}
@@ Line 74: / Line 79: @@
 !Step 1: &nbsp;
 |-
-|We need to use the limit definition of derivative and calculate the limit from both sides.
+|We need to use the limit definition of derivative and calculate the limit from both sides. So, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\
 &&\\
@@ Line 96: / Line 99: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4\sqrt{3+h+1}-8}{h}}\\
 &&\\
@@ Line 118: / Line 121: @@
 !Step 3: &nbsp;
 |-
-|Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
+|Since
 |-
-|<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
+|<math style="vertical-align: -5px">f(x)</math>&nbsp; is differentiable at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous.
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>&thinsp; is continuous at <math style="vertical-align: 0px">x=3.</math>
 |-
-|&nbsp;&nbsp; '''(b)''' Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
 |-
+|
-::<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable at <math style="vertical-align: 0px">x=3.</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]