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| | <span class="exam"> Consider the following piecewise defined function: | | <span class="exam"> Consider the following piecewise defined function: |
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| − | ::::::<math>f(x) = \left\{
| + | ::<math>f(x) = \left\{ |
| | \begin{array}{lr} | | \begin{array}{lr} |
| | x+5 & \text{if }x < 3\\ | | x+5 & \text{if }x < 3\\ |
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| | \right. | | \right. |
| | </math> | | </math> |
| − | <span class="exam">a) Show that <math style="vertical-align:-10%">f(x)</math> is continuous at <math style="vertical-align:0%">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:-10%">f(x)</math> is differentiable at <math style="vertical-align:0%">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>''']] |
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| − | |} | |
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009A Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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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: -12px">\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} 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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| − | {| 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>\lim_{x\rightarrow 3^-}f(x)</math>. We have
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| − | |-
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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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| − | {| 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>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 <math>\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3)</math>, <math>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;"
| |
| − | !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 <math>\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>f(x)</math> is differentiable at <math>x=3</math>.
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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>\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3)</math>, <math>f(x)</math> is continuous.
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| − | |-
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| − | |'''(b)''' Since <math>\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>f(x)</math> is differentiable at <math>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>''']] |
