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| − | <span class="exam">Consider the following function <math style="vertical-align: -5px"> f:</math> | + | <span class="exam">Suppose the size of a population at time <math style="vertical-align: 0px">t</math> is given by |
| − | ::<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 <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math> | + | ::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</math> |
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| − | <span class="exam">(b) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math> | + | <span class="exam">(a) Determine the size of the population as <math style="vertical-align: -1px">t\rightarrow \infty.</math> We call this the limiting population size. |
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| − | <span class="exam">(c) Find <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math> | + | <span class="exam">(b) Show that at time <math style="vertical-align: -4px">t=5,</math> the size of the population is half its limiting size. |
| | + | <hr> |
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| − | <span class="exam">(d) Is <math style="vertical-align: -5px">f</math> continuous at <math style="vertical-align: -1px">x=1?</math> Briefly explain.
| + | (insert picture of handwritten solution) |
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| | + | [[009A Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''1.''' If <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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| − | | then <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
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| − | |-
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| − | |'''2.''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=a</math> if
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| − | |-
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| − | | <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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| − |
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| − | '''Solution:'''
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| − |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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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 <math style="vertical-align: 0px">x</math> that are smaller than <math style="vertical-align: -1px">1.</math>
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| − | |-
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| − | |Using the definition of <math style="vertical-align: -5px">f(x),</math> we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
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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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| − | !Step 2:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | <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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| − |
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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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 <math style="vertical-align: 0px">x</math> that are bigger than <math style="vertical-align: -2px">1.</math>
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| − | |-
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| − | |Using the definition of <math style="vertical-align: -5px">f(x),</math> we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
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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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| − | !Step 2:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | <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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| − |
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |From (a) and (b), we have
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| − | |-
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| − | | <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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| − | | <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
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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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| − | !Step 2:
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| − | |-
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| − | |Since
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| − | |-
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| − | | <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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| − | | <math>\lim_{x\rightarrow 1}f(x)=1.</math>
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| − | |}
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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:
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| − | |-
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| − | |From (c), we have
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| − | |-
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| − | | <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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| − | | <math>f(1)=\sqrt{1}=1.</math>
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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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| − | !Step 2:
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| − | |-
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| − | |Since
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| − | |-
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| − | | <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> is continuous at <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:
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| − | |-
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| − | | '''(a)''' <math>1</math>
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| − | |-
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| − | | '''(b)''' <math>1</math>
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| − | |-
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| − | | '''(c)''' <math>1</math>
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| − | |-
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| − | | '''(d)''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1</math> since <math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Suppose the size of a population 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}
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 
the size of the population is half its limiting size.
(insert picture of handwritten solution)
Detailed Solution for this Problem
Return to Sample Exam