# 009A Sample Final A, Problem 8

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8. (a) Find the linear approximation ${\displaystyle L(x)}$ to the function ${\displaystyle f(x)=\sec x}$ at the point ${\displaystyle x=\pi /3}$.
(b) Use ${\displaystyle L(x)}$ to estimate the value of ${\displaystyle \sec \,(3\pi /7)}$.

Foundations:
Recall that the linear approximation ${\displaystyle L(x)}$ is the equation of the tangent line to a function at a given point. If we are given the point ${\displaystyle x_{0}}$, then we will have the approximation ${\displaystyle L(x)=f'(x_{0})\cdot (x-x_{0})+f(x_{0})}$. Note that such an approximation is usually only good "fairly close" to your original point ${\displaystyle x_{0}}$.

Solution:

Part (a):
Note that ${\displaystyle f'(x)=\sec x\tan x}$. Since ${\displaystyle \sin(\pi /3)={\sqrt {3}}/2}$ and ${\displaystyle \cos(\pi /3)=1/2}$, we have
${\displaystyle f'(\pi /3)\,\,=\,\,\sec(\pi /3)\tan(\pi /3)\,\,=\,\,{\frac {1}{1/2}}\cdot {\frac {{\sqrt {3}}/2}{\,\,1/2}}\,\,=\,\,2{\sqrt {3}}.}$
Similarly, ${\displaystyle f(\pi /3)=\sec(\pi /3)=2.}$ Together, this means that
${\displaystyle L(x)=f'(x_{0})\cdot (x-x_{0})+f(x_{0})}$
${\displaystyle =2{\sqrt {3}}(x-\pi /3)+2.}$
Part (b):
This is simply an exercise in plugging in values. We have

${\displaystyle L\left({\frac {3\pi }{7}}\right)=2{\sqrt {3}}\left({\frac {3\pi }{7}}-{\frac {\pi }{3}}\right)+2}$
${\displaystyle =2{\sqrt {3}}\left({\frac {9\pi -7\pi }{21}}\right)+2}$
${\displaystyle =2{\sqrt {3}}\left({\frac {2\pi }{21}}\right)+2}$
${\displaystyle ={\frac {4{\sqrt {3}}\pi }{21}}+2.}$