Difference between revisions of "009A Sample Final 1, Problem 2"

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!Step 1:    
 
!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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|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)</math>. We have
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we calculate <math style="vertical-align: -12px">\lim_{x\rightarrow 3^-}f(x)</math>. We have
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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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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|Now, we calculate <math style="vertical-align:-10%">f(3)</math>. We have
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|Now, we calculate <math style="vertical-align: -3px">f(3)</math>. We have
 
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|<math>f(3)=4\sqrt{3+1}=8</math>.
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::<math>f(3)=4\sqrt{3+1}=8</math>.
 
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|Since <math style="vertical-align: -12px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=f(3)</math>, <math style="vertical-align: -2px">f(x)</math> is continuous.
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|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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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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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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|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>f(x)</math> is differentiable at <math>x=3</math>.
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|<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">x=3</math>.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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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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|'''(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>\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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|'''(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>f(x)</math> is differentiable at <math>x=3</math>.
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|
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::<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">x=3</math>.
 
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|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:56, 22 February 2016

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:  

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} .

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