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

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(Created page with "<span class="exam">Consider the following function  <math style="vertical-align: -5px"> f:</math> ::<math>f(x) = \left\{ \begin{array}{lr} x^2 & \text{if }x...")
 
 
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<span class="exam">Consider the following function &nbsp;<math style="vertical-align: -5px"> f:</math>
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<span class="exam">Suppose the size of a population at time &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is given by
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::<math>f(x) = \left\{
 
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    \begin{array}{lr}
 
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      x^2 & \text{if }x < 1\\
 
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      \sqrt{x} & \text{if }x \geq 1
 
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    \end{array}
 
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  \right.
 
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</math>
 
  
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<span class="exam">(a) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
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::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</math>
  
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<span class="exam">(b) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
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<span class="exam">(a) Determine the size of the population as &nbsp;<math style="vertical-align: -1px">t\rightarrow \infty.</math>&nbsp; We call this the limiting population size.
  
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<span class="exam">(c) Find &nbsp;<math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>
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<span class="exam">(b) Show that at time &nbsp;<math style="vertical-align: -4px">t=5,</math>&nbsp; the size of the population is half its limiting size.
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<hr>
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[[009A Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']]
  
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<span class="exam">(d) Is &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; continuous at &nbsp;<math style="vertical-align: -1px">x=1?</math>&nbsp; Briefly explain.
 
  
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[[009A Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Foundations: &nbsp;
 
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|-
 
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|'''1.''' If &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
 
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|-
 
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|'''2.''' &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=a</math>&nbsp; if
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
 
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|}
 
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'''Solution:'''
 
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'''(a)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|Notice that we are calculating a left hand limit.
 
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|-
 
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|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are smaller than &nbsp;<math style="vertical-align: -1px">1.</math>
 
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|-
 
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|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Now, we have
 
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|-
 
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|
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow 1^-} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^-} x^2}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{x\rightarrow 1} x^2}\\
 
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&&\\
 
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& = & \displaystyle{1^2}\\
 
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&&\\
 
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& = & \displaystyle{1.}\\
 
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\end{array}</math>
 
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|}
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|Notice that we are calculating a right hand limit.
 
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|-
 
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|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are bigger than &nbsp;<math style="vertical-align: -2px">1.</math>
 
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|-
 
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|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Now, we have
 
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|-
 
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|
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow 1^+} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^+} \sqrt{x}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{x\rightarrow 1} \sqrt{x}}\\
 
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&&\\
 
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& = & \displaystyle{\sqrt{1}}\\
 
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&&\\
 
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& = & \displaystyle{1.}\\
 
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\end{array}</math>
 
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|}
 
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'''(c)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|From (a) and (b), we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f(x)=1</math>
 
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|-
 
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|and
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Since
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}f(x)=1,</math>
 
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|-
 
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|we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=1.</math>
 
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|}
 
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'''(d)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|From (c), we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=1.</math>
 
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|-
 
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|Also,
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)=\sqrt{1}=1.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Since
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
 
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|-
 
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|<math style="vertical-align: -5px">f(x)</math> &nbsp;is continuous at &nbsp;<math style="vertical-align: -1px">x=1.</math>
 
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|-
 
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|
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|-
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>1</math>
 
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|-
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>1</math>
 
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|-
 
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>1</math>
 
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|-
 
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|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: -1px">x=1</math>&nbsp; since &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
 
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|}
 
 
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 14:40, 8 November 2017

Suppose the size of a population at time    is given by

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 N(t)=\frac{1000t}{5+t},~t\ge 0.}

(a) Determine the size of the population as  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 t\rightarrow \infty.}   We call this the limiting population size.

(b) Show that at time  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 t=5,}   the size of the population is half its limiting size.

Solution


Detailed Solution


Return to Sample Exam

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