The linear approximation $L ( x )$ to $f ( x ) = \sqrt { x }$ near $x _ { 0 } = \frac { 1 } { 4 }$ is

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

Question

The linear approximation $L ( x )$ to $f ( x ) = \sqrt { x }$ near $x _ { 0 } = \frac { 1 } { 4 }$ is

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

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

The Linear Approximation L(x)L ( x )L(x) To f(x)=xf ( x ) = \sqrt { x }f(x)=x​

The Linear Approximation $L ( x )$ To $f ( x ) = \sqrt { x }$