Difference between revisions of "009A Sample Final A, Problem 8"

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! Foundations:    
 
! Foundations:    
 
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|Recall that the linear approximation ''L''(''x'') is the equation of the tangent line to a function at a given point. If we are given the point ''x''<span style="font-size:85%"><sub>0</sub></span>, then we will have the approximation <math style="vertical-align: -20%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>.  Note that such an approximation is usually only good "fairly close" to your original point  ''x''<span style="font-size:85%"><sub>0</sub></span>.
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|Recall that the linear approximation <math style="vertical-align: -25%;">L(x)</math> is the equation of the tangent line to a function at a given point. If we are given the point <math style="vertical-align: -15%;">x_0</math>, then we will have the approximation <math style="vertical-align: -20%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>.  Note that such an approximation is usually only good "fairly close" to your original point  <math style="vertical-align: -15%;">x_0</math>.
 
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&nbsp;'''Solution:'''
 
&nbsp;'''Solution:'''

Revision as of 21:11, 26 March 2015


8. (a) Find the linear approximation to the function at the point .
    (b) Use to estimate the value of .

Foundations:  
Recall that the linear approximation is the equation of the tangent line to a function at a given point. If we are given the point , then we will have the approximation . Note that such an approximation is usually only good "fairly close" to your original point .

 Solution:

Part (a):  
Note that f '(x) = sec x tan x. Since sin(π/3) = √3/2 and cos(π/3) = 1/2, we have
    
Similarly, f(π/3) = sec(π/3) = 2. Together, this means that
    
                
Part (b):  
This is simply an exercise in plugging in values. We have

    
                        
                        
                        

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