009A Sample Final 1, Problem 9 - Revision history

Kayla Murray at 22:59, 2 December 2017

2017-12-02T22:59:41Z

Kayla Murray at 20:24, 18 March 2017

2017-03-18T20:24:52Z

Kayla Murray at 16:38, 27 February 2017

2017-02-27T16:38:04Z

Kayla Murray at 01:06, 26 February 2017

2017-02-26T01:06:07Z

Kayla Murray at 01:49, 19 February 2017

2017-02-19T01:49:34Z

Kayla Murray at 01:45, 19 February 2017

2017-02-19T01:45:08Z

Grad at 21:08, 4 March 2016

2016-03-04T21:08:34Z

Grad: /* 7 */

2016-03-04T21:07:56Z

Grad: /* 5 */

2016-03-04T21:06:06Z

Grad: /* 6 */

2016-03-04T21:05:19Z

@@ Line 11: / Line 11: @@
 <span class="exam">(e) Use the above information (a) to (d) to sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x)</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 9 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 33: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp; We have &nbsp;<math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
+|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 |-
-|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<math style="vertical-align: -6px">0=3x(x-4).</math>
+|We have &nbsp;<math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
 |-
 |Hence, we have &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=4.</math>
 |-
-|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
+|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 |}
@@ Line 81: / Line 87: @@
 |We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |-
-|So, we have &nbsp;<math style="vertical-align: -1px">0=6x-12.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=2.</math>
+|So, we have
 |-
-|This value breaks up the number line into two intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
 |}

@@ 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">Given the function &nbsp;<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.
@@ Line 9: / Line 9: @@
 <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 &nbsp;<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>&nbsp; is increasing when <math style="vertical-align: -5px">f'(x)>0</math>&nbsp; and <math style="vertical-align: -5px">f(x)</math>&nbsp; is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
+|'''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>
 |-
 |'''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>&nbsp; is concave up when <math style="vertical-align: -5px">f''(x)>0</math>&nbsp; and <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
+|'''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>
 |-
-|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
+|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 |}
@@ Line 33: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>&nbsp; We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
+|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp; We have &nbsp;<math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
 |-
-|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
+|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<math style="vertical-align: -6px">0=3x(x-4).</math>
 |-
-|Hence, we have <math style="vertical-align: 0px">x=0</math>&nbsp; and <math style="vertical-align: -1px">x=4.</math>
+|Hence, we have &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<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: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
+|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 |}
@@ Line 47: / Line 47: @@
 |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 &nbsp;<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 &nbsp;<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>
+|For &nbsp;<math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
 |-
-|Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>&nbsp; and decreasing on <math style="vertical-align: -5px">(0,4).</math>
+|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(0,4).</math>
 |}
@@ Line 61: / Line 61: @@
 !Step 1: &nbsp;
 |-
-|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>
+|By the First Derivative Test, the local maximum occurs at &nbsp;<math style="vertical-align: -1px">x=0</math>&nbsp; and the local minimum occurs at &nbsp;<math style="vertical-align: -1px">x=4.</math>
 |}
@@ Line 67: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|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>
+|So, the local maximum value is &nbsp;<math style="vertical-align: -5px">f(0)=5</math>&nbsp; and the local minimum value is &nbsp;<math style="vertical-align: -5px">f(4)=-27.</math>
 |}
@@ Line 75: / Line 75: @@
 !Step 1: &nbsp;
 |-
-|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>
+|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>
 |-
-|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
+|We have &nbsp;<math style="vertical-align: -5px">f''(x)=6x-12.</math>
 |-
-|We set <math style="vertical-align: -5px">f''(x)=0.</math>
+|We set &nbsp;<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 &nbsp;<math style="vertical-align: -1px">0=6x-12.</math>&nbsp; Hence, &nbsp;<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>
+|This value breaks up the number line into two intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
 |}
@@ Line 91: / Line 91: @@
 |Again, we use test points in these two intervals.
 |-
-|For <math style="vertical-align: -5px">x=0,</math>&nbsp; we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
+|For &nbsp;<math style="vertical-align: -5px">x=0,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-12<0.</math>
 |-
-|For <math style="vertical-align: -5px">x=3,</math>&nbsp; we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
+|For &nbsp;<math style="vertical-align: -5px">x=3,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=6>0.</math>
 |-
-|Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; 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>
+|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(2,\infty),</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,2).</math>
 |}
@@ Line 101: / Line 101: @@
 !(d) &nbsp;
 |-
-| Using the information from part (c), there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
+| Using the information from part (c), there is one inflection point that occurs at &nbsp;<math style="vertical-align: 0px">x=2.</math>
 |-
-|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
+|Now, we have &nbsp;<math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
 |-
 |So, the inflection point is &nbsp;<math style="vertical-align: -5px">(2,-11).</math>
@@ Line 118: / Line 118: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; 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)''' &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-\infty,0),(4,\infty),</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(0,4).</math>
 |-
-|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>&nbsp; and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;The local maximum value is &nbsp;<math style="vertical-align: -5px">f(0)=5,</math>&nbsp; and the local minimum value is &nbsp;<math style="vertical-align: -5px">f(4)=-27.</math>
 |-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; 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)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(2,\infty),</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,2).</math>
 |-
-|&nbsp; &nbsp;'''(d)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(2,-11)</math>
+|&nbsp; &nbsp;'''(d)''' &nbsp; &nbsp; <math style="vertical-align: -5px">(2,-11)</math>
 |-
-|&nbsp; &nbsp;'''(e)''' &nbsp; &nbsp;See graph above.
+|&nbsp; &nbsp;'''(e)''' &nbsp; &nbsp; See graph above.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ 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>
+|'''1.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing when <math style="vertical-align: -5px">f'(x)>0</math>&nbsp; and <math style="vertical-align: -5px">f(x)</math>&nbsp; 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>
+|'''3.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up when <math style="vertical-align: -5px">f''(x)>0</math>&nbsp; and <math style="vertical-align: -5px">f(x)</math>&nbsp; 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 33: / Line 33: @@
 !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>&nbsp; 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>&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>
+|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<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>
+|Hence, we have <math style="vertical-align: 0px">x=0</math>&nbsp; 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: &thinsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
+|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 |}
@@ Line 53: / Line 53: @@
 |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>
+|Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>&nbsp; and decreasing on <math style="vertical-align: -5px">(0,4).</math>
 |}
@@ Line 91: / Line 91: @@
 |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=0,</math>&nbsp; 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>
+|For <math style="vertical-align: -5px">x=3,</math>&nbsp; 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>
+|Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; 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>
 |}
@@ Line 101: / Line 101: @@
 !(d) &nbsp;
 |-
-| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
+| 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 &thinsp;<math style="vertical-align: -5px">(2,-11).</math>
+|So, the inflection point is &nbsp;<math style="vertical-align: -5px">(2,-11).</math>
 |}
@@ Line 118: / Line 118: @@
 !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)''' &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; 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)''' &nbsp; &nbsp;The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>&nbsp; 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)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; 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)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(2,-11)</math>
 |-
-|'''(e)''' See graph in '''(e)'''.
+|&nbsp; &nbsp;'''(e)''' &nbsp; &nbsp;See graph above.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 24: / Line 24: @@
 |'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
 |}
 '''Solution:'''
@@ Line 112: / Line 113: @@
 | Insert sketch here.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ 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;"