Difference between revisions of "009A Sample Final 1, Problem 2"

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<span class="exam"> Consider the following piecewise defined function:
 
<span class="exam"> Consider the following piecewise defined function:
  
−
::::::<math>f(x) = \left\{
+
::<math>f(x) = \left\{
 
     \begin{array}{lr}
 
     \begin{array}{lr}
 
       x+5 &  \text{if }x < 3\\
 
       x+5 &  \text{if }x < 3\\
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   \right.
 
   \right.
 
</math>
 
</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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!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>
+
|-
 +
|'''2.''' The definition of derivative for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
−
::'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math>&thinsp; is &thinsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
 
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!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}}\\
 
\displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\
 
&&\\
 
&&\\
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!Step 2: &nbsp;
 
!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}\\
 
\displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\
 
&&\\
 
&&\\
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!Step 3: &nbsp;
 
!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  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
 +
|-
 +
|<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!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
−
|-
 
−
|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}}\\
 
\displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
::<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}}\\
 
\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}}\\
 
&&\\
 
&&\\
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!Step 3: &nbsp;
 
!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  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<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>
+
|<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!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>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 08:07, 10 April 2017

Consider the following piecewise defined function:

(a) Show that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is continuous at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.}

(b) Using the limit definition of the derivative, and computing the limits from both sides, show that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is differentiable at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} .

Foundations:  
1.  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is continuous at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a}   if
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).}
2. The definition of derivative for  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.}


Solution:

(a)

Step 1:  
We first calculate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x).}   We have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 2:  
Now, we calculate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^-}f(x).}   We have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\ &&\\ & = & \displaystyle{3+5}\\ &&\\ & = & \displaystyle{8.} \end{array}}

Step 3:  
Now, we calculate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(3).}   We have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(3)=4\sqrt{3+1}\,=\,8.}

Since
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is continuous at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.}

(b)

Step 1:  
We need to use the limit definition of derivative and calculate the limit from both sides. So, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 2:  
Now, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 3:  
Since
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is differentiable at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.}


Final Answer:  
    (a)     Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)}   is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.}
    (b)     Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},}

              Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.}

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