Difference between revisions of "009A Sample Final A, Problem 8"
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− | |Recall that the linear approximation <math style="vertical-align: - | + | |Recall that the linear approximation <math style="vertical-align: -22%;">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: -12%;">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: -12%;">x_0</math>. |
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'''Solution:''' | '''Solution:''' | ||
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− | |Note that | + | |Note that <math style="vertical-align: -17%;">f'(x) = \sec x \tan x</math>. Since <math style="vertical-align: -20%;">\sin(\pi/3)=\sqrt{3}/2</math> and <math style="vertical-align: -20%;">\cos(\pi/3)=1/2</math>, we have |
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| <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. </math> | | <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. </math> | ||
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− | |Similarly, | + | |Similarly, <math style="vertical-align: -22%;">f(\pi/3) = \sec(\pi/3) = 2</math>. Together, this means that |
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| <math>L(x) = f'(x_0)\cdot (x-x_0)+f(x_0) </math> | | <math>L(x) = f'(x_0)\cdot (x-x_0)+f(x_0) </math> |
Revision as of 21:19, 26 March 2015
8. (a) Find the linear approximation to the function at the point .
(b) Use to estimate the value of .
Foundations: |
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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): |
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Note that . Since and , we have |
Similarly, . Together, this means that |
Part (b): |
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This is simply an exercise in plugging in values. We have |