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| | <span class="exam"> For what values of <math style="vertical-align: 0px">x</math> is <math style="vertical-align: -4px">f</math> continuous? | | <span class="exam"> For what values of <math style="vertical-align: 0px">x</math> is <math style="vertical-align: -4px">f</math> continuous? |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 2, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=a</math> if
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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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| − | '''Solution:''' | + | [[009A Sample Final 2, 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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| − | |Based on the description of <math style="vertical-align: -5px">f(x),</math>
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| − | |-
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| − | |we know <math style="vertical-align: -5px">f(x)</math> is continuous on
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| − | |-
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| − | | <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</math>
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| − | |-
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| − | |Now, we need to see if <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=3.</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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| − | |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^+} \frac{x^2-2x-3}{x-3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 3^+} \frac{(x-3)(x+1)}{x-3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 3^+} x+1}\\
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| − | &&\\
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| − | & = & \displaystyle{4.}
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| − | \end{array}</math>
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| − | |-
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| − | |Similarly,
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| − | | <math>\lim_{x\rightarrow 3^-}f(x)=4.</math>
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| − | |-
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| − | |Since
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| − | |-
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| − | | <math style="vertical-align: 0px">\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow 3^+}f(x)</math>
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 3}f(x)=4.</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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| − | |But, since
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| − | | <math>f(3)=5,</math>
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 3}f(x)\ne f(3).</math>
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| − | |-
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| − | |Therefore, <math style="vertical-align: -5px">f(x)</math> is not continuous at <math style="vertical-align: 0px">x=3.</math>
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| − | |-
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| − | | <math style="vertical-align: -5px">f(x)</math> is continuous only on <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</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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| − | | <math style="vertical-align: -5px">f(x)</math> is continuous on <math style="vertical-align: -5px">(-\infty,3)\cup (3,\infty).</math>
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| − | |}
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| | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
