|
|
| (3 intermediate revisions by the same user not shown) |
| Line 9: |
Line 9: |
| | | | |
| | <span class="exam">(d) Sketch the graph of <math style="vertical-align: -5px">y=f(x).</math> | | <span class="exam">(d) Sketch the graph of <math style="vertical-align: -5px">y=f(x).</math> |
| | + | <hr> |
| | + | [[009A Sample Final 2, Problem 10 Solution|'''<u>Solution</u>''']] |
| | | | |
| − | {| 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>
| |
| − | |}
| |
| | | | |
| | + | [[009A Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''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>
| |
| − | |-
| |
| − | |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:
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | '''(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)'''
| |
| − | |-
| |
| − | | '''(c)''' <math>y=0</math>
| |
| − | |-
| |
| − | | '''(d)''' See above
| |
| − | |}
| |
| | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
