MathAdmin at 19:55, 26 May 2017

2017-05-26T19:55:37Z

MathAdmin at 01:13, 21 May 2017

2017-05-21T01:13:04Z

MathAdmin: Created page with "Let :: $f(x)=\frac{4x}{x^2+1}$ (a) Find all local maximum and local minimum values of
2017-04-10T16:26:40Z

Created page with "Let ::<math>f(x)=\frac{4x}{x^2+1}</math> (a) Find all local maximum and local minimum values of <math style="vertical-align: -4px"..."

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Let
::<math>f(x)=\frac{4x}{x^2+1}</math>

(a) Find all local maximum and local minimum values of  <math style="vertical-align: -4px">f,</math>  find all intervals where  <math style="vertical-align: -4px">f</math>  is increasing and all intervals where  <math style="vertical-align: -4px">f</math>  is decreasing.

(b) Find all inflection points of the function  <math style="vertical-align: -4px">f,</math>  find all intervals where the function  <math style="vertical-align: -4px">f</math>  is concave upward and all intervals where  <math style="vertical-align: -4px">f</math>  is concave downward.

(c) Find all horizontal asymptotes of the graph  <math style="vertical-align: -5px">y=f(x).</math>

(d) Sketch the graph of  <math style="vertical-align: -5px">y=f(x).</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when  <math style="vertical-align: -5px">f'(x)>0</math>  and  <math style="vertical-align: -5px">f(x)</math>  is decreasing when  <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when  <math style="vertical-align: -5px">f''(x)>0</math>  and  <math style="vertical-align: -5px">f(x)</math>  is concave down when  <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when  <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of  <math style="vertical-align: -5px">f(x).</math> 
|-
|Using the Quotient Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{(x^2+1)(4x)'-(4x)(x^2+1)'}{(x^2+1)^2}}\\
&&\\
& = & \displaystyle{\frac{(x^2+1)(4)-(4x)(2x)}{(x^2+1)^2}}\\
&&\\
& = & \displaystyle{\frac{-4(x^2-1)}{(x^2+1)^2}.}
\end{array}</math>
|-
|Now, we set  <math style="vertical-align: -5px">f'(x)=0.</math> 
|-
|So, we have
|-
|       <math style="vertical-align: -6px">-4(x^2-1)=0.</math>
|-
|Hence, we have  <math style="vertical-align: 0px">x=1</math>  and  <math style="vertical-align: -1px">x=-1.</math>
|-
|So, these values of  <math style="vertical-align: 0px">x</math>  break up the number line into 3 intervals:
|-
|       <math style="vertical-align: -5px">(-\infty,-1),(-1,1),(1,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For  <math style="vertical-align: -15px">x=-2,~f'(x)=\frac{-12}{25}<0.</math>
|-
|For  <math style="vertical-align: -5px">x=0,~f'(x)=4>0.</math>
|-
|For  <math style="vertical-align: -15px">x=2,~f'(x)=\frac{-12}{25}<0.</math>
|-
|Thus,  <math style="vertical-align: -5px">f(x)</math>  is increasing on  <math style="vertical-align: -5px">(-1,1)</math>  and decreasing on  <math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Using the First Derivative Test,  <math style="vertical-align: -5px">f(x)</math>  has a local minimum at  <math style="vertical-align: -1px">x=-1</math>  and a local maximum at  <math style="vertical-align: 0px">x=1.</math> 
|-
|Thus, the local maximum and local minimum values of  <math style="vertical-align: -4px">f</math>  are
|-
|       <math>f(1)=2,~f(-1)=-2.</math>
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find  <math style="vertical-align: -5px">f''(x).</math>
|-
|Using the Quotient Rule and Chain Rule, we have
|-
|        <math>\begin{array}{rcl}
\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}}\\
&&\\
& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(x^2+1)'}{(x^2+1)^4}}\\
&&\\
& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(2x)}{(x^2+1)^4}}\\
&&\\
& = & \displaystyle{\frac{(x^2+1)(-8x)+16(x^2-1)x}{(x^2+1)^3}}\\
&&\\
& = & \displaystyle{\frac{8x^3-24x}{(x^2+1)^3}}\\
&&\\
& = & \displaystyle{\frac{8x(x^2-3)}{(x^2+1)^3}.}
\end{array}</math>
|-
|We set  <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have
|-
|       <math style="vertical-align: -1px">8x(x^2-3)=0.</math> 
|-
|Hence,
|-
|       <math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
|-
|This value breaks up the number line into four intervals:
|-
|       <math style="vertical-align: -5px">(-\infty,-\sqrt{3}),(-\sqrt{3},0),(0,\sqrt{3}),(\sqrt{3},\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these four intervals.
|-
|For  <math style="vertical-align: -5px">x=-2,</math>  we have  <math style="vertical-align: -15px">f''(x)=\frac{-16}{125}<0.</math>
|-
|For  <math style="vertical-align: -5px">x=-1,</math>  we have  <math style="vertical-align: -5px">f''(x)=2>0.</math>
|-
|For  <math style="vertical-align: -5px">x=1,</math>  we have  <math style="vertical-align: -5px">f''(x)=-2<0.</math>
|-
|For  <math style="vertical-align: -5px">x=2,</math>  we have  <math style="vertical-align: -15px">f''(x)=\frac{16}{125}>0.</math>
|-
|Thus,  <math style="vertical-align: -5px">f(x)</math>  is concave up on  <math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>  and concave down on  <math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|The inflection points occur at
|-
|       <math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
|-
|Plugging these into  <math style="vertical-align: -5px">f(x),</math> we get the inflection points
|-
|       <math style="vertical-align: 0px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that the degree of the numerator is  <math style="vertical-align: -1px">1</math>  and
|-
|the degree of the denominator is  <math style="vertical-align: 0px">2.</math> 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the degree of the denominator is greater than the degree of the numerator,
|-
|<math style="vertical-align: -5px">f(x)</math>  has a horizontal asymptote
|-
|       <math>y=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d):  
|-
|Insert sketch
|-
|
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math style="vertical-align: -5px">f(x)</math>  is increasing on  <math style="vertical-align: -5px">(-1,1)</math>  and decreasing on  <math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
|-
|           The local maximum value of  <math style="vertical-align: -4px">f</math>  is  <math style="vertical-align: 0px">2</math>  and the local minimum value of  <math style="vertical-align: -4px">f</math>  is  <math style="vertical-align: 0px">-2</math> 
|-
|   '''(b)'''    <math style="vertical-align: -5px">f(x)</math>  is concave up on  <math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>  and concave down on  <math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
|-
|            The inflection points are  <math style="vertical-align: -5px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</math>
|-
|   '''(c)'''    <math>y=0</math>
|-
|   '''(d)'''    See above
|}
[[009A_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 188: / Line 188: @@
 !(d): &nbsp;
 |-
-|Insert sketch
+|[[File:009A_SF2_10.jpg|left|720px]]
 |-
+|

@@ Line 152: / Line 152: @@
 !Step 1: &nbsp;
 |-
-|First, we note that the degree of the numerator is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and
+|By L'Hopital's Rule, we have
 |-
-|the degree of the denominator is &nbsp;<math style="vertical-align: 0px">2.</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 160: / Line 176: @@
 !Step 2: &nbsp;
 |-
-|Since the degree of the denominator is greater than the degree of the numerator,
+|Since
 |-
 |<math style="vertical-align: -5px">f(x)</math>&nbsp; has a horizontal asymptote

009A Sample Final 2, Problem 10 - Revision history

MathAdmin at 19:55, 26 May 2017

MathAdmin at 01:13, 21 May 2017