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| − | <span class="exam">(a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3</math>. | + | <span class="exam">(a) Show that <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3.</math> |
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| | <span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>. | | <span class="exam">(b) Using the limit definition of the derivative, and computing the limits from both sides, show that <math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3</math>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']] |
| − | |- | |
| − | |'''1.''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=a</math> if
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| − | |-
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| − | | <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
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| − | |-
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| − | |'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math> is
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| − | |-
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| − | | <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009A Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math> We have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{4\sqrt{3+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{8.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math> We have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\
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| − | &&\\
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| − | & = & \displaystyle{3+5}\\
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| − | &&\\
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| − | & = & \displaystyle{8.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, we calculate <math style="vertical-align: -5px">f(3).</math> We have
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| − | |-
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| − | <math>f(3)=4\sqrt{3+1}\,=\,8.</math>
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| − | |-
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| − | |Since
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| − | |-
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| − | | <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),</math>
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| − | |-
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| − | |<math style="vertical-align: -5px">f(x)</math> is continuous.
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We need to use the limit definition of derivative and calculate the limit from both sides. So, we have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{h}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^-}1}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4\sqrt{3+h+1}-8}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})(\sqrt{4+h}+\sqrt{4})}{h(\sqrt{4+h}+\sqrt{4})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(4+h-4)}{h(\sqrt{4+h}+\sqrt{4})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4h}{h(\sqrt{4+h}+\sqrt{4})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4}{(\sqrt{4+h}+\sqrt{4})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{2\sqrt{4}}}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}\\
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Since
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| − | |-
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| − | | <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
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| − | |-
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| − | |<math style="vertical-align: -5px">f(x)</math> is differentiable at <math style="vertical-align: 0px">x=3.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' Since <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>  is continuous.
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| − | |-
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| − | | '''(b)''' Since <math style="vertical-align: -14px">\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},</math>
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| − | |-
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| − | |
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| − | <math style="vertical-align: -5px">f(x)</math>  is differentiable at <math style="vertical-align: 0px">x=3.</math>
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
