Difference between revisions of "009A Sample Final 1, Problem 2"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
| − | |||
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|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is continuous at <math style="vertical-align: 0px">x=a</math>  if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math> | |'''1.''' <math style="vertical-align: -5px">f(x)</math>  is continuous at <math style="vertical-align: 0px">x=a</math>  if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math> | ||
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| − | + | <math>\begin{array}{rcl} | |
\displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\ | \displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\ | ||
&&\\ | &&\\ | ||
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| − | + | <math>\begin{array}{rcl} | |
\displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\ | \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\ | ||
&&\\ | &&\\ | ||
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| − | + | <math>f(3)=4\sqrt{3+1}\,=\,8.</math> | |
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|Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>  is continuous. | |Since <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)</math>  is continuous. | ||
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| − | + | <math>\begin{array}{rcl} | |
\displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\ | \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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| − | + | <math>\begin{array}{rcl} | |
\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}}\\ | \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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!Final Answer: | !Final Answer: | ||
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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. | + | | '''(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. |
|- | |- | ||
| − | |'''(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> | + | | '''(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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| − | + | <math style="vertical-align: -5px">f(x)</math>  is differentiable at <math style="vertical-align: 0px">x=3.</math> | |
|} | |} | ||
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 16:26, 25 February 2017
Consider the following piecewise defined function:
(a) Show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} .
(b) Using the limit definition of the derivative, and computing the limits from both sides, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} .
| Foundations: |
|---|
| 1. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a} if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).} |
| 2. The definition of derivative for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.} |
Solution:
(a)
| Step 1: |
|---|
| We first calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x).} We have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow 3^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^+} 4\sqrt{x+1}}\\ &&\\ & = & \displaystyle{4\sqrt{3+1}}\\ &&\\ & = & \displaystyle{8.} \end{array}} |
| Step 2: |
|---|
| Now, we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^-}f(x).} We have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow 3^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 3^-} x+5}\\ &&\\ & = & \displaystyle{3+5}\\ &&\\ & = & \displaystyle{8.} \end{array}} |
| Step 3: |
|---|
| Now, we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(3).} We have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(3)=4\sqrt{3+1}\,=\,8.} |
| Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)} is continuous. |
(b)
| Step 1: |
|---|
| We need to use the limit definition of derivative and calculate the limit from both sides. So, we have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}} & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{(3+h)+5-8}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^-}\frac{h}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^-}1}\\ &&\\ & = & \displaystyle{1.} \end{array}} |
| Step 2: |
|---|
| Now, we have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \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}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(\sqrt{4+h}-\sqrt{4})(\sqrt{4+h}+\sqrt{4})}{h(\sqrt{4+h}+\sqrt{4})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4(4+h-4)}{h(\sqrt{4+h}+\sqrt{4})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4h}{h(\sqrt{4+h}+\sqrt{4})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0^+}\frac{4}{(\sqrt{4+h}+\sqrt{4})}}\\ &&\\ & = & \displaystyle{\frac{4}{2\sqrt{4}}}\\ &&\\ & = & \displaystyle{1.}\\ \end{array}} |
| Step 3: |
|---|
| Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},} |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.} |
| Final Answer: |
|---|
| (a) Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3),~f(x)} is continuous. |
| (b) Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},} |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3.} |