Difference between revisions of "009A Sample Final 2, Problem 10"

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<span class="exam">(d) Sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
 
<span class="exam">(d) Sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
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[[009A Sample Final 2, Problem 10 Solution|'''<u>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.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing when &nbsp;<math style="vertical-align: -5px">f'(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is decreasing when &nbsp;<math style="vertical-align: -5px">f'(x)<0.</math>
 
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|-
 
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|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
 
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|'''3.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up when &nbsp;<math style="vertical-align: -5px">f''(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave down when &nbsp;<math style="vertical-align: -5px">f''(x)<0.</math>
 
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|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|}
 
  
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[[009A Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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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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|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 
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|Using the Quotient Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f'(x)} & = & \displaystyle{\frac{(x^2+1)(4x)'-(4x)(x^2+1)'}{(x^2+1)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)(4)-(4x)(2x)}{(x^2+1)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{-4(x^2-1)}{(x^2+1)^2}.}
 
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\end{array}</math>
 
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|-
 
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|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp;
 
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|-
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -6px">-4(x^2-1)=0.</math>
 
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|Hence, we have &nbsp;<math style="vertical-align: 0px">x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 
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|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-1),(-1,1),(1,\infty).</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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|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 
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|-
 
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|For &nbsp;<math style="vertical-align: -15px">x=-2,~f'(x)=\frac{-12}{25}<0.</math>
 
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|-
 
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|For &nbsp;<math style="vertical-align: -5px">x=0,~f'(x)=4>0.</math>
 
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|For &nbsp;<math style="vertical-align: -15px">x=2,~f'(x)=\frac{-12}{25}<0.</math>
 
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|-
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-1,1)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Using the First Derivative Test, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has a local minimum at &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; and a local maximum at &nbsp;<math style="vertical-align: 0px">x=1.</math>&nbsp;
 
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|-
 
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|Thus, the local maximum and local minimum values of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; are
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(1)=2,~f(-1)=-2.</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: &nbsp;
 
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|-
 
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|To find the intervals when the function is concave up or concave down, we need to find &nbsp;<math style="vertical-align: -5px">f''(x).</math>
 
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|Using the Quotient Rule and Chain Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f''(x)} & = & \displaystyle{\frac{(x^2+1)^2(-4(x^2-1))'+4(x^2-1)((x^2+1)^2)'}{((x^2+1)^2)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(x^2+1)'}{(x^2+1)^4}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(2x)}{(x^2+1)^4}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)(-8x)+16(x^2-1)x}{(x^2+1)^3}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{8x^3-24x}{(x^2+1)^3}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{8x(x^2-3)}{(x^2+1)^3}.}
 
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\end{array}</math>
 
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|-
 
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|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">8x(x^2-3)=0.</math>&nbsp;
 
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|Hence,
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
 
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|This value breaks up the number line into four intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3}),(-\sqrt{3},0),(0,\sqrt{3}),(\sqrt{3},\infty).</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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|Again, we use test points in these four intervals.
 
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|For &nbsp;<math style="vertical-align: -5px">x=-2,</math>&nbsp; we have &nbsp;<math style="vertical-align: -15px">f''(x)=\frac{-16}{125}<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=2>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-2<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=2,</math>&nbsp; we have &nbsp;<math style="vertical-align: -15px">f''(x)=\frac{16}{125}>0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on &nbsp;<math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>&nbsp; and concave down on &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|The inflection points occur at
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
 
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|Plugging these into &nbsp;<math style="vertical-align: -5px">f(x),</math> we get the inflection points
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</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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|First, we note that the degree of the numerator is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and
 
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|the degree of the denominator is &nbsp;<math style="vertical-align: 0px">2.</math>&nbsp;
 
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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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|Since the degree of the denominator is greater than the degree of the numerator,
 
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|<math style="vertical-align: -5px">f(x)</math>&nbsp; has a horizontal asymptote
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=0.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(d): &nbsp;
 
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|Insert sketch
 
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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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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-1,1)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;The local maximum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">2</math>&nbsp; and the local minimum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">-2</math>&nbsp;
 
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on &nbsp;<math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>&nbsp; and concave down on &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; The inflection points are &nbsp;<math style="vertical-align: -5px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</math>
 
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>y=0</math>
 
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|&nbsp; &nbsp;'''(d)'''&nbsp; &nbsp; See above
 
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|}
 
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 10:29, 1 December 2017

Let

(a) Find all local maximum and local minimum values 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,}   find all intervals where  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 increasing and all intervals where  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 decreasing.

(b) Find all inflection points of the 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,}   find all intervals where the 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 concave upward and all intervals where  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 concave downward.

(c) Find all horizontal asymptotes of the graph  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 y=f(x).}

(d) Sketch the graph 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 y=f(x).}

Solution


Detailed Solution


Return to Sample Exam

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