Difference between revisions of "009A Sample Final A, Problem 3"

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(Created page with "<span style="font-size:135%"><font face=Times Roman>3. (Version I) Consider the following function:  <math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, &...")
 
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<span style="font-size:135%"><font face=Times Roman>3. (Version I) Consider the following function:
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[[File:009A_SF_A_3.png|right|230px|frame|Both functions with constants chosen to provide continuity.]]
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<span class="exam">3. (Version I) Consider the following function:
 
&nbsp;<math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, & \mbox{if }x\geq 1, \\ 4x^{2}+C, & \mbox{if }x<1. \end{cases}</math>
 
&nbsp;<math style="vertical-align: -80%;">f(x) = \begin{cases} \sqrt{x}, & \mbox{if }x\geq 1, \\ 4x^{2}+C, & \mbox{if }x<1. \end{cases}</math>
 
<br>
 
<br>
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&nbsp;&nbsp; (a) Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -3%;">x=1.</math>  
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<table border="0" class="exam">
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<br>
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<tr style="vertical-align:top">
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&nbsp;&nbsp; (b) With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer.
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<td>&nbsp;(a) </td>
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<br><br>  
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<td>Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -3%;">x=1.</math></td>  
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3. (Version II) Consider the following function:
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<tr style="vertical-align:top">
 +
<td>&nbsp;(b)</td>
 +
<td>With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer.</td>
 +
</table>  
 +
 
 +
<span class="exam">3. (Version II) Consider the following function:
 
&nbsp;<math style="vertical-align: -80%;">g(x)=\begin{cases}
 
&nbsp;<math style="vertical-align: -80%;">g(x)=\begin{cases}
 
\sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\
 
\sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\
 
\frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1.
 
\frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1.
 
\end{cases}</math>
 
\end{cases}</math>
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<br>
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<table border="0" class="exam">
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&nbsp;&nbsp; (a) Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -2.95%;">x=1.</math>  
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<tr style="vertical-align:top">
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<br>
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<td>&nbsp;(a) </td>
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&nbsp;&nbsp; (b) With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer. </font face=Times Roman> </span>
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<td>Find a value of &nbsp;<math style="vertical-align: -0.1%;">C</math> which makes <math>f</math> continuous at <math style="vertical-align: -2.95%;">x=1.</math>  
 +
</td>
 +
<tr style="vertical-align:top">
 +
<td>&nbsp;(b)</td>
 +
<td>With your choice of &nbsp;<math style="vertical-align: -0.1%;">C</math>, is <math>f</math> differentiable at <math style="vertical-align: -3%;">x=1</math>? &nbsp;Use the definition of the derivative to motivate your answer.</td>
 +
</table>
 +
 
 +
<font face="arial,helvetica">
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
! Foundations: &nbsp;  
 
! Foundations: &nbsp;  
 
|-
 
|-
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|A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -15%;">x_0 </math> if  
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|A function <math style="vertical-align: -20%;">f</math> is continuous at a point <math style="vertical-align: -14%">x_0 </math> if  
 
|-
 
|-
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_0} f(x) = f\left(x_0\right).</math>
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow x_{_0}} f(x) = f\left(x_0\right).</math>
 
|-
 
|-
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|This can be viewed as saying the left and right hand limits exist, and are equal. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
+
|This can be viewed as saying the left and right hand limits exist, and are equal to the value of <math style="vertical-align: -20%">f</math> at <math style="vertical-align: -14%">x_0</math>. For problems like these, where we are trying to find a particular value for&thinsp; <math style="vertical-align: 0%;">C</math>, we can just set the two descriptions of the function to be equal at the value where the definition of <math style="vertical-align: -20%;">f</math> changes.
 
|-
 
|-
−
|When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledge that the derivative is a limit, and for a limit to exist it must agree from the left and the right.  This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br>
+
|When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right.  This means we must show the derivatives agree for both the descriptions of <math style="vertical-align: -20%;">f</math> at the transition point.<br>
 
|}
 
|}
 +
 +
&nbsp;'''Solution:'''
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
! Version I: &nbsp;  
 
! Version I: &nbsp;  
 
|-
 
|-
 +
|(a) For continuity, we evaluate both rules for the function at the transition point <math style="vertical-align: -3%;">x=1</math>, set the results equal, and then solve for <math style="vertical-align: -2%;">C</math>.  Since we want
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;  <math>f(1)\,=\,1\,=\,4\cdot 1^2+C,</math>
 +
|-
 +
|we can set <math style="vertical-align: 0%;">C=-3</math>, and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point <math style="vertical-align: -2%;">x=1</math>&thinsp;).
  
 +
(b) To test differentiability, we note that for <math style="vertical-align: -2%;">x>1</math>,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{1}{2\sqrt{x}},</math>
 +
|-
 +
|while for <math style="vertical-align: -3%;">x<1</math>,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=8x.</math>
 +
|-
 +
|Thus
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+}f'(x)\,=\,\frac{1}{2\sqrt{1}}\,=\,\frac{1}{2},</math>
 +
|-
 +
|but
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,8\cdot1\,=\,8.</math>
 +
|-
 +
|Since the left and right hand limit do not agree, the derivative does not exist at the point <math style="vertical-align: -2%;">x=1</math>.<br>
 
|}
 
|}
  
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! Version II: &nbsp;  
 
! Version II: &nbsp;  
 
|-
 
|-
 +
|(a) Like Version I, we begin by setting the two functions equal.  We want
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,</math>
 +
|-
 +
|so <math style="vertical-align: -24%;">C=7/4</math> makes the function continuous.
 +
|-
 +
|(b) We again consider the derivative from each side of 1. For <math style="vertical-align: -2%;">x>1</math>,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{1}{2\sqrt{x^2+3}}\cdot 2x\,=\,\frac{x}{\sqrt{x^2+3}},</math>
 +
|-
 +
|while for <math style="vertical-align: -3%;">x<1</math>,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)\,=\,\frac{x}{2}.</math>
 +
|-
 +
|Thus
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+}f'(x)\,=\,\frac{1}{\sqrt{1^2+3}}\,=\,\frac{1}{2},</math>
 +
|-
 +
|and
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.</math>
 +
|-
 +
|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>, and <math style="vertical-align: -15%;">g(x)</math><br> is differentiable at the required point.
 +
|}
  
−
|}
 
  
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:33, 12 April 2015

Both functions with constants chosen to provide continuity.

3. (Version I) Consider the following function:  

 (a) Find a value of   which makes 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} 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=1.}
 (b) With your choice of  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 C} , 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} 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=1} ?  Use the definition of the derivative to motivate your answer.

3. (Version II) Consider the following function:  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 g(x)=\begin{cases} \sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\ \frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1. \end{cases}}

 (a) Find a value of  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 C} which makes 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} 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=1.}
 (b) With your choice of  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 C} , 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} 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=1} ?  Use the definition of the derivative to motivate your answer.

Foundations:  
A function 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} is continuous at a point 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_0 } 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 x_{_0}} f(x) = f\left(x_0\right).}
This can be viewed as saying the left and right hand limits exist, and are equal to the value of 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} 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_0} . For problems like these, where we are trying to find a particular value 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 C} , we can just set the two descriptions of the function to be equal at the value where the definition of 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} changes.
When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledging that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of 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} at the transition point.

 Solution:

Version I:  
(a) For continuity, we evaluate both rules for the function at the transition point 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=1} , set the results equal, and then solve 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 C} . Since we want
        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(1)\,=\,1\,=\,4\cdot 1^2+C,}
we can set 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 C=-3} , and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point 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=1}  ).

(b) To test differentiability, we note that 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 x>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)=\frac{1}{2\sqrt{x}},}
while 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 x<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)=8x.}
Thus
        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 1^+}f'(x)\,=\,\frac{1}{2\sqrt{1}}\,=\,\frac{1}{2},}
but
        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 1^-}f'(x)\,=\,8\cdot1\,=\,8.}
Since the left and right hand limit do not agree, the derivative does not exist at the point 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=1} .
Version II:  
(a) Like Version I, we begin by setting the two functions equal. We want
        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(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,}
so 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 C=7/4} makes the function continuous.
(b) We again consider the derivative from each side of 1. 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 x>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)=\frac{1}{2\sqrt{x^2+3}}\cdot 2x\,=\,\frac{x}{\sqrt{x^2+3}},}
while 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 x<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)\,=\,\frac{x}{2}.}
Thus
        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 1^+}f'(x)\,=\,\frac{1}{\sqrt{1^2+3}}\,=\,\frac{1}{2},}
and
        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 1^-}f'(x)\,=\,\frac{1}{2}.}
Since the left and right hand limit do agree, the limit (which is the derivative) does exist at the point 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=1} , and 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 g(x)}
is differentiable at the required point.


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