Find the linear approximation to $f ( x ) = \frac { 1 } { ( 2 + x ) ^ { 3 } }$ at $a = 0$
A)
$\frac { 1 } { 8 } - \frac { 1 } { 8 } x$
B)
$\frac { 1 } { 8 } + \frac { 1 } { 8 } x$
C)
$\frac { 1 } { 8 } + \frac { 3 } { 16 } x$
D)
$\frac { 1 } { 8 } + \frac { 1 } { 16 } x$
E) $\frac { 1 } { 8 } - \frac { 3 } { 16 } x$
F) $\frac { 1 } { 8 } - \frac { 3 } { 4 } x$
G) $\frac { 1 } { 8 } - \frac { 1 } { 16 } x$
H)
$\frac { 1 } { 8 } + \frac { 3 } { 4 } x$

Question

Find the linear approximation to $f ( x ) = \frac { 1 } { ( 2 + x ) ^ { 3 } }$ at $a = 0$
A)
$\frac { 1 } { 8 } - \frac { 1 } { 8 } x$
B)
$\frac { 1 } { 8 } + \frac { 1 } { 8 } x$ 
C)
$\frac { 1 } { 8 } + \frac { 3 } { 16 } x$
D)
$\frac { 1 } { 8 } + \frac { 1 } { 16 } x$
E) $\frac { 1 } { 8 } - \frac { 3 } { 16 } x$ 
F) $\frac { 1 } { 8 } - \frac { 3 } { 4 } x$
G) $\frac { 1 } { 8 } - \frac { 1 } { 16 } x$  
H)
$\frac { 1 } { 8 } + \frac { 3 } { 4 } x$

Quizplus · Accepted Answer

The linear approximation of a function $f(x)$ at $x = a$ is given by $f(a) + f'(a)(x-a)$. Here, $f(x) = \frac{1}{(2+x)^3}$, so $f(0) = \frac{1}{8}$. The derivative $f'(x) = -\frac{3}{(2+x)^4}$, so $f'(0) = -\frac{3}{16}$. Thus, the linear approximation is $\frac{1}{8} - \frac{3}{16}x$.

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Find the Linear Approximation To $f ( x ) = \frac { 1 } { ( 2 + x ) ^ { 3 } }$