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

From Grad Wiki
Jump to navigation Jump to search
Line 17: Line 17:
 
|Recall:
 
|Recall:
 
|-
 
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">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: -1px">x=a</math> if <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
 
|-
 
|-
|'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>.
+
|'''2.''' The definition of derivative for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.</math>
 
|}
 
|}
  
Line 29: Line 29:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x)</math>. We have
+
|We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^+}f(x).</math> We have
 
|-
 
|-
 
|
 
|
Line 37: Line 37:
 
& = & \displaystyle{4\sqrt{3+1}}\\
 
& = & \displaystyle{4\sqrt{3+1}}\\
 
&&\\
 
&&\\
& = & \displaystyle{8}
+
& = & \displaystyle{8.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 44: Line 44:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x)</math>. We have
+
|Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 3^-}f(x).</math> We have
 
|-
 
|-
 
|
 
|
Line 52: Line 52:
 
& = & \displaystyle{3+5}\\
 
& = & \displaystyle{3+5}\\
 
&&\\
 
&&\\
& = & \displaystyle{8}
+
& = & \displaystyle{8.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 59: Line 59:
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
|Now, we calculate <math style="vertical-align: -3px">f(3)</math>. We have
+
|Now, we calculate <math style="vertical-align: -3px">f(3).</math> We have
 
|-
 
|-
 
|
 
|
::<math>f(3)=4\sqrt{3+1}=8</math>.
+
::<math>f(3)=4\sqrt{3+1}=8.</math>
 
|-
 
|-
 
|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.
 
|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.
Line 82: Line 82:
 
& = & \displaystyle{\lim_{h\rightarrow 0^-}1}\\
 
& = & \displaystyle{\lim_{h\rightarrow 0^-}1}\\
 
&&\\
 
&&\\
& = & \displaystyle{1}
+
& = & \displaystyle{1.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 107: Line 107:
 
& = & \displaystyle{\frac{4}{2\sqrt{4}}}\\
 
& = & \displaystyle{\frac{4}{2\sqrt{4}}}\\
 
&&\\
 
&&\\
& = & \displaystyle{1}\\
+
& = & \displaystyle{1.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 114: Line 114:
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
|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>,
+
|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>  
 
|-
 
|-
|<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">x=3</math>.
+
|<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">x=3.</math>
 
|}
 
|}
  
Line 124: Line 124:
 
|'''(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>  
 
|-
 
|-
 
|
 
|
::<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">x=3</math>.
+
::<math style="vertical-align: -3px">f(x)</math> is differentiable at <math style="vertical-align: -1px">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 12:27, 29 February 2016

Consider the following piecewise defined function:

a) Show that is continuous at .

b) Using the limit definition of the derivative, and computing the limits from both sides, show that is differentiable at .

Foundations:  
Recall:
1. is continuous at if
2. The definition of derivative for is

Solution:

(a)

Step 1:  
We first calculate We have
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.}

Return to Sample Exam