022 Exam 2 Sample A, Problem 9 - Revision history

MathAdmin at 15:12, 16 May 2015

2015-05-16T15:12:01Z

MathAdmin at 05:41, 16 May 2015

2015-05-16T05:41:59Z

MathAdmin at 05:38, 16 May 2015

2015-05-16T05:38:03Z

MathAdmin: Created page with " Find all relative extrema and points of inflection for the function $g(x) = \frac{2}{3}x^3 + x^2 - 12x$ . Be sure to..."

2015-05-16T05:29:56Z

Created page with "<span class="exam"> Find all relative extrema and points of inflection for the function <math style="vertical-align: -45%">g(x) = \frac{2}{3}x^3 + x^2 - 12x</math>. Be sure to..."

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<span class="exam">
Find all relative extrema and points of inflection for the function <math style="vertical-align: -45%">g(x) = \frac{2}{3}x^3 + x^2 - 12x</math>. Be sure to give coordinate pairs for each point. You do not need to draw the graph.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Since our function is a polynomial, the relative extrema occur when the first derivative is zero. We then have two choices for finding if it is a local maximum or minimum:
|-
|'''Second Derivative Test:''' If the first derivative at a point <math style="vertical-align: -12%">x_0</math> is <math style="vertical-align: 0%">0</math>, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local maximum.
|-
|On the other hand, if the second derivative is positive, the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local minimum. You can also use the first derivative test, but it is usually a bit more work! For inflection points, we need to find when the second derivative is zero, as well as check that the second derivative "splits" on both sides.
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''Find the first and second derivatives:''' Based on our function, we have
|-
|
::<math>g\,'(x)\,=\,\frac{2}{3}\cdot 3x^2+2x-12\,=\,2x^2+2x-12.</math>
|-
|Similarly, from the first derivative we find
|-
|
::<math>g\,''(x)\,=\,4x+2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''Find the roots of the derivatives:''' We can rewrite the first derivative as
|-
|
::<math>g\,'(x)\,=\,2x^2+2x-12\,=\,2(x^2+x-6)\,=\,2(x+3)(x-2),</math>
|-
|from which it should be clear we have roots <math style="vertical-align: 0%">2</math> and <math style="vertical-align: 0%">-3</math>.
|-
|On the other hand, for the second derivative, we have
|-
|
::<math>g\,''(x)\,=\,4x+2\,=\,2\left(x+\frac{1}{2}\right).</math>
|-
|This has a single root: <math style="vertical-align: -60%">x=-\frac{1}{2}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|'''Test the potential extrema:''' We know that <math>x=2,-3</math> are the candidates. We check the second derivative, finding
|-
|
::<math>g\,''(2)\,=\,4\cdot 2+2\,>\,0,</math>
|-
|while
|-
|
::<math>g\,''(-3)\,=\,2(-3)+2\,<\,0.</math>
|-
|Note that
|-
|
::<math>g(2)\,=\,\frac{2}{3}(8)+4-24\,=\,-\frac{44}{3},</math>
|-
|while
|-
|
::<math>g(-3)\,=\,\frac{2}{3}(-27)+9-12(-3)\,=\,27.</math>
|-
|By the second derivative test, the point <math style="vertical-align: -70%">(2,g(2))=\left(2,-\frac{44}{3}\right)</math> is a relative maximum, while the point <math style="vertical-align: -22%">(-3,g(-3))=(-3,27)</math> is a relative maximum.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|'''Test the potential inflection point:''' We know that <math style="vertical-align: -70%">g\,''\left(-\frac{1}{2}\right)=0</math>. On the other hand, it should be clear that if <math style="vertical-align: -60%">x<-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)<0</math>. Similarly, if <math style="vertical-align: -60%">x>-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)>0</math>. Thus, the second derivative "splits" around <math style="vertical-align: -60%">x=-\frac{1}{2}</math>  (i.e., changes sign), so the point <math style="vertical-align: -70%">\left(-\frac{1}{2},g\left(-\frac{1}{2}\right) \right)</math>  is an inflection point.
|-
|Since
|-
|
::<math>g\left(-\frac{1}{2}\right)\,=\,\frac{2}{3}\cdot-\frac{1}{8}+\frac{1}{4}-12\left(-\frac{1}{2}\right)\,=\,\frac{19}{4},</math>
|-
|our inflection point is
|-
|
::<math>\left(-\frac{1}{2},\frac{19}{4}\right).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|The area is maximized when both the length and width are 12 meters.
|}

[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 |'''Second Derivative Test:''' If the first derivative at a point <math style="vertical-align: -12%">x_0</math> is <math style="vertical-align: 0%">0</math>, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local maximum.
 |-
-|On the other hand, if the second derivative is positive, the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local minimum.  You can also use the first derivative test, but it is usually a bit more work!  For inflection points, we need to find when the second derivative is zero, as well as check that the second derivative "splits" on both sides.
+|On the other hand, if the second derivative is positive, the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local minimum.  You can also use the first derivative test, but it is usually a bit more work!  For '''inflection points''', we need to find when the second derivative is zero, as well as check that the second derivative "splits" on both sides.
 |}
@@ Line 41: / Line 41: @@
 |-
+|
-::<math>g\,''(x)\,=\,4x+2\,=\,2\left(x+\frac{1}{2}\right).</math>
+::<math>g\,''(x)\,=\,4x+2\,=\,4\left(x+\frac{1}{2}\right).</math>
 |-
 |This has a single root: <math style="vertical-align: -60%">x=-\frac{1}{2}</math>.

@@ Line 80: / Line 80: @@
 |-
+|
-::<math>g\left(-\frac{1}{2}\right)\,=\,\frac{2}{3}\cdot-\frac{1}{8}+\frac{1}{4}-12\left(-\frac{1}{2}\right)\,=\,\frac{19}{4},</math>
+::<math>g\left(-\frac{1}{2}\right)\,=\,\frac{2}{3}\left(-\frac{1}{8}\right)+\frac{1}{4}-12\left(-\frac{1}{2}\right)\,=\,\frac{37}{6},</math>
 |-
 |our inflection point is
 |-
+|
-::<math>\left(-\frac{1}{2},\frac{19}{4}\right).</math>
+::<math>\left(-\frac{1}{2},\frac{37}{6}\right).</math>
 |}
@@ Line 91: / Line 91: @@
 !Final Answer: &nbsp;
 |-
-|There is a local minimum at <math style="vertical-align: -70%">\left(2,-\frac{44}{3}\right)</math>, a local maximum at <math style="vertical-align: -22%">(-3,27)</math> and an inflection point at <math style="vertical-align: -70%">\left(-\frac{1}{2},\frac{19}{4}\right).</math>
+|There is a local minimum at <math style="vertical-align: -70%">\left(2,-\frac{44}{3}\right)</math>, a local maximum at <math style="vertical-align: -22%">(-3,27)</math> and an inflection point at <math style="vertical-align: -70%">\left(-\frac{1}{2},\frac{37}{6}\right).</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 ::<math>g(-3)\,=\,\frac{2}{3}(-27)+9-12(-3)\,=\,27.</math>
 |-
-|By the second derivative test, the point <math style="vertical-align: -70%">(2,g(2))=\left(2,-\frac{44}{3}\right)</math> is a relative maximum, while the point <math style="vertical-align: -22%">(-3,g(-3))=(-3,27)</math> is a relative maximum.
+|By the second derivative test, the point <math style="vertical-align: -70%">(2,g(2))=\left(2,-\frac{44}{3}\right)</math> is a relative minimum, while the point <math style="vertical-align: -22%">(-3,g(-3))=(-3,27)</math> is a relative maximum.
 |}
@@ Line 75: / Line 75: @@
 !Step 4: &nbsp;
 |-
-|'''Test the potential inflection point:''' We know that <math style="vertical-align: -70%">g\,''\left(-\frac{1}{2}\right)=0</math>. On the other hand, it should be clear that if <math style="vertical-align: -60%">x<-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)<0</math>.  Similarly, if <math style="vertical-align: -60%">x>-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)>0</math>. Thus, the second derivative "splits" around <math style="vertical-align: -60%">x=-\frac{1}{2}</math>&thinsp; (i.e., changes sign), so the point <math style="vertical-align: -70%">\left(-\frac{1}{2},g\left(-\frac{1}{2}\right) \right)</math>&thinsp; is an inflection point.
+|'''Test the potential inflection point:''' We know that <math style="vertical-align: -25%">g\,''(-1/2)=0</math>. On the other hand, it should be clear that if <math style="vertical-align: -25%">x<-1/2</math>, then <math style="vertical-align: -23%">g\,''(x)<0</math>.  Similarly, if <math style="vertical-align: -25%">x>-1/2</math>, then <math style="vertical-align: -23%">g\,''(x)>0</math>. Thus, the second derivative "splits" around <math style="vertical-align: -25%">x=-1/2</math>&thinsp; (i.e., changes sign), so the point <math style="vertical-align: -25%">\left(-1/2,g(-1/2)\right)</math>&thinsp; is an inflection point.
 |-
 |Since
@@ Line 91: / Line 91: @@
 !Final Answer: &nbsp;
 |-
-|The area is maximized when both the length and width are 12 meters.
+|There is a local minimum at <math style="vertical-align: -70%">\left(2,-\frac{44}{3}\right)</math>, a local maximum at <math style="vertical-align: -22%">(-3,27)</math> and an inflection point at <math style="vertical-align: -70%">\left(-\frac{1}{2},\frac{19}{4}\right).</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

022 Exam 2 Sample A, Problem 9 - Revision history

MathAdmin at 15:12, 16 May 2015

MathAdmin at 05:41, 16 May 2015

MathAdmin at 05:38, 16 May 2015

MathAdmin: Created page with " Find all relative extrema and points of inflection for the function g(x) = \frac{2}{3}x^3 + x^2 - 12x. Be sure to..."

MathAdmin: Created page with " Find all relative extrema and points of inflection for the function $g(x) = \frac{2}{3}x^3 + x^2 - 12x$ . Be sure to..."