The linear approximation $L ( x )$ to $f ( x ) = \tan x$ near $x _ { 0 } = \frac { \pi } { 4 }$ is

A) $1 + 2 \left( x - \frac { \pi } { 4 } \right)$
B) $1 - 2 \left( x - \frac { \pi } { 4 } \right)$
C) $1 + \left( x - \frac { \pi } { 4 } \right)$
D) $1 - \left( x - \frac { \pi } { 4 } \right)$
E) $- 1 + 2 \left( x - \frac { \pi } { 4 } \right)$

Question

The linear approximation $L ( x )$ to $f ( x ) = \tan x$ near $x _ { 0 } = \frac { \pi } { 4 }$ is

A) $1 + 2 \left( x - \frac { \pi } { 4 } \right)$
B) $1 - 2 \left( x - \frac { \pi } { 4 } \right)$
C) $1 + \left( x - \frac { \pi } { 4 } \right)$
D) $1 - \left( x - \frac { \pi } { 4 } \right)$
E) $- 1 + 2 \left( x - \frac { \pi } { 4 } \right)$

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 $f(x) = \tan(x)$ at $x_0 = \frac{\pi}{4}$, $f(x_0) = \tan(\frac{\pi}{4}) = 1$ and $f'(x) = \sec^2(x)$, so $f'(x_0) = \sec^2(\frac{\pi}{4}) = 2$. Thus, $L(x) = 1 + 2(x - \frac{\pi}{4})$.

The Linear Approximation L(x)L ( x )L(x) To f(x)=tan⁡xf ( x ) = \tan xf(x)=tanx

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