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

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|&nbsp; &nbsp; &nbsp; &nbsp;  <math>f(1)\,=\,1\,=\,4\cdot 1^2+C,</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;  <math>f(1)\,=\,1\,=\,4\cdot 1^2+C,</math>
 
|-
 
|-
|we can set <math style="vertical-align: 1%;">C=-3</math>, and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point <math style="vertical-align: -2%;">x=1</math>&thinsp;).
+
|we can set <math style="vertical-align: 0%;">C=-3</math>, and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point <math style="vertical-align: -2%;">x=1</math>&thinsp;).
  
 
(b) To test differentiability, we note that for <math style="vertical-align: -2%;">x<1</math>,
 
(b) To test differentiability, we note that for <math style="vertical-align: -2%;">x<1</math>,
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{1}{2\sqrt{x}},</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{1}{2\sqrt{x}},</math>
 
|-
 
|-
|while for <math style="vertical-align: -5%;">x> 1</math>,
+
|while for <math style="vertical-align: -3%;">x> 1</math>,
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=8x.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=8x.</math>

Revision as of 13:27, 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:  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 g(x)=\begin{cases} \sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\ \frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1. \end{cases}}
   (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:  

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