009A Sample Final 1, Problem 9 - Revision history

MathAdmin at 16:17, 10 April 2017

2017-04-10T16:17:28Z

MathAdmin at 16:16, 10 April 2017

2017-04-10T16:16:26Z

Show changes

MathAdmin at 22:16, 18 April 2016

2016-04-18T22:16:28Z

MathAdmin at 19:23, 18 April 2016

2016-04-18T19:23:43Z

MathAdmin at 07:21, 5 March 2016

2016-03-05T07:21:32Z

MathAdmin: Created page with "Given the function $f(x)=x^3-6x^2+5$ , a) Find the intervals in which the function increases or..."

2016-03-05T07:11:09Z

Created page with "Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>, a) Find the intervals in which the function increases or..."

New page

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''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>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</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,0),(0,4),(4,\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: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}

'''(c)'''

{| 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>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d)  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

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

{| 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">(-\infty,0),(4,\infty),</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>  and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See graph in '''(e)'''.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 123: / Line 123: @@
 |[[File: 9AF1_9_GP.png|700px|center]]
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 110: / Line 110: @@
 !(e) &nbsp;
 |-
-| Insert sketch here.
+|[[File: 9AF1_9_GP.png|700px|center]]
 |}

@@ Line 137: / Line 137: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty),</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
+|&nbsp;&nbsp; '''(a)''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty),</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
 |-
-|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>&thinsp; and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
+|&nbsp;&nbsp; '''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>&thinsp; and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
 |-
-|'''(c)''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
+|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
 |-
-|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
+|&nbsp;&nbsp; '''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
 |-
-|'''(e)''' See graph in '''(e)'''.
+|&nbsp;&nbsp; '''(e)''' See graph in '''(e)'''.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,
-<span class="exam">a) Find the intervals in which the function increases or decreases.
+::<span class="exam">a) Find the intervals in which the function increases or decreases.
-<span class="exam">b) Find the local maximum and local minimum values.
+::<span class="exam">b) Find the local maximum and local minimum values.
-<span class="exam">c) Find the intervals in which the function concaves upward or concaves downward.
+::<span class="exam">c) Find the intervals in which the function concaves upward or concaves downward.
-<span class="exam">d) Find the inflection point(s).
+::<span class="exam">d) Find the inflection point(s).
-<span class="exam">e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.
+::<span class="exam">e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 16: / Line 16: @@
 |Recall:
 |-
 |'''1.''' <math style="vertical-align: -5px">f(x)</math>&thinsp; is increasing when <math style="vertical-align: -5px">f'(x)>0</math>&thinsp; and <math style="vertical-align: -5px">f(x)</math>&thinsp; 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>&thinsp; is concave up when <math style="vertical-align: -5px">f''(x)>0</math>&thinsp; and <math style="vertical-align: -5px">f(x)</math>&thinsp; 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>
 |}
@@ Line 32: / Line 36: @@
 !Step 1: &nbsp;
 |-
-|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>&thinsp; We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
+|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>&thinsp; We have
 |-
-|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>&thinsp; So, we have &thinsp;<math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
 |-
 |Hence, we have <math style="vertical-align: 0px">x=0</math>&thinsp; and <math style="vertical-align: -1px">x=4.</math>
@@ Line 46: / Line 56: @@
 |To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 |-
 |For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
 |-
 |For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
 |-
 |For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
 |-
 |Thus, <math style="vertical-align: -5px">f(x)</math>&thinsp; is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>&thinsp; and decreasing on <math style="vertical-align: -5px">(0,4).</math>
@@ Line 76: / Line 89: @@
 |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>
 |-
-|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
+|We have
 |-
 |We set <math style="vertical-align: -5px">f''(x)=0.</math>
 |-
-|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
+|So, we have
 |-
-|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
+|Hence, <math style="vertical-align: 0px">x=2.</math> This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
 |}
@@ Line 90: / Line 109: @@
 |Again, we use test points in these two intervals.
 |-
 |For <math style="vertical-align: -5px">x=0,</math>&thinsp; we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
 |-
 |For <math style="vertical-align: -5px">x=3,</math>&thinsp; we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
 |-
 |Thus, <math style="vertical-align: -5px">f(x)</math>&thinsp; is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>