The linear approximation $L ( x )$ to $f ( x ) = \sin x$ near $x _ { 0 } = - \frac { \pi } { 6 }$ is

A) $- \frac { 1 } { 2 } - \frac { \sqrt { 3 } } { 2 } \left( x + \frac { \pi } { 6 } \right)$
B) $- \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } \left( x + \frac { \pi } { 6 } \right)$
C) $- \frac { \sqrt { 3 } } { 2 } + \frac { 1 } { 2 } \left( x + \frac { \pi } { 6 } \right)$
D) $- \frac { \sqrt { 3 } } { 2 } - \frac { 1 } { 2 } \left( x + \frac { \pi } { 6 } \right)$
E) $- \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } \left( x - \frac { \pi } { 6 } \right)$

Question

The linear approximation $L ( x )$ to $f ( x ) = \sin x$ near $x _ { 0 } = - \frac { \pi } { 6 }$ is

A) $- \frac { 1 } { 2 } - \frac { \sqrt { 3 } } { 2 } \left( x + \frac { \pi } { 6 } \right)$
B) $- \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } \left( x + \frac { \pi } { 6 } \right)$
C) $- \frac { \sqrt { 3 } } { 2 } + \frac { 1 } { 2 } \left( x + \frac { \pi } { 6 } \right)$
D) $- \frac { \sqrt { 3 } } { 2 } - \frac { 1 } { 2 } \left( x + \frac { \pi } { 6 } \right)$
E) $- \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } \left( x - \frac { \pi } { 6 } \right)$

Quizplus · Accepted Answer

The linear approximation of a function $f(x)$ at a point $x_0$ is given by $L(x) = f(x_0) + f'(x_0)(x - x_0)$. For $f(x) = \sin x$ at $x_0 = -\frac{\pi}{6}$, $f(x_0) = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$ and $f'(x) = \cos x$, so $f'(x_0) = \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. Thus, $L(x) = -\frac{1}{2} + \frac{\sqrt{3}}{2}(x + \frac{\pi}{6})$.

The Linear Approximation L(x)L ( x )L(x) To f(x)=sin⁡xf ( x ) = \sin xf(x)=sinx

The Linear Approximation $L ( x )$ To $f ( x ) = \sin x$