Difference between revisions of "009A Sample Final A, Problem 3"

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! Version II:    
 
! Version II:    
 
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|-
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|(a) Like Version I, we begin by setting the two functions equal.  We want
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,</math>
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|-
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|so <math style="vertical-align: -24%;">C=-7/4</math> makes the function continuous.
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|-
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|(b) We again consider the derivative from each side of 1. For <math style="vertical-align: -2%;">x<1</math>,
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{1}{2\sqrt{x^2+3}}\cdot 2x\,=\,\frac{x}{\sqrt{x^2+3}},</math>
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|-
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|while for <math style="vertical-align: -3%;">x> 1</math>,
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)\,=\,\frac{x}{2}.</math>
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|-
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|Thus
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{\sqrt{1^2+3}}\,=\,\frac{1}{2},</math>
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|-
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|and
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+}f'(x)\,=\,\frac{1}{2}.</math>
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|-
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|Since the left and right hand limit do agree, the limit (which <u>''is''</u> the derivative) does exist at the point <math style="vertical-align: -2%;">x=1</math>.<br>
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[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:38, 27 March 2015

3. (Version I) Consider the following function:  
   (a) Find a value of   which makes continuous at
   (b) With your choice of  , is differentiable at ?  Use the definition of the derivative to motivate your answer.

3. (Version II) Consider the following function:  
   (a) Find a value of   which makes continuous at
   (b) With your choice of  , is differentiable at ?  Use the definition of the derivative to motivate your answer.

Foundations:  
A function is continuous at a point if
    
This can be viewed as saying the left and right hand limits exist, and are equal. For problems like these, where we are trying to find a particular value for  , we can just set the two descriptions of the function to be equal at the value where the definition of changes.
When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledge that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of at the transition point.

 Solution:

Version I:  
(a) For continuity, we evaluate both rules for the function at the transition point , set the results equal, and then solve for . Since we want
       
we can set , and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point  ).

(b) To test differentiability, we note that for ,

       
while for ,
       
Thus
       
but
       
Since the left and right hand limit do not agree, the derivative does not exist at the point .
Version II:  
(a) Like Version I, we begin by setting the two functions equal. We want
       
so makes the function continuous.
(b) We again consider the derivative from each side of 1. For ,
       
while for ,
       
Thus
       
and
       
Since the left and right hand limit do agree, the limit (which is the derivative) does exist at the point .


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