Find the linear approximation to $f ( x ) = \frac { 1 } { \sqrt { x ^ { 3 } + 1 } }$ at $a = 2 .$
A)-2x + 7
B) $\frac { 1 } { 3 } x - \frac { 2 } { 3 }$
C) $- \frac { 2 } { 9 } x + \frac { 7 } { 9 }$
D) $\frac { 1 } { 3 } x - \frac { 7 } { 3 }$
E) $2 x - 7$
F)
$- \frac { 1 } { 3 } x + \frac { 2 } { 3 }$
G)
$\frac { 2 } { 9 } x - \frac { 7 } { 9 }$
H) $- \frac { 1 } { 3 } x + \frac { 7 } { 3 }$

Question

Find the linear approximation to $f ( x ) = \frac { 1 } { \sqrt { x ^ { 3 } + 1 } }$ at $a = 2 .$
A)-2x + 7
B) $\frac { 1 } { 3 } x - \frac { 2 } { 3 }$
C) $- \frac { 2 } { 9 } x + \frac { 7 } { 9 }$
D) $\frac { 1 } { 3 } x - \frac { 7 } { 3 }$
E) $2 x - 7$
F)
$- \frac { 1 } { 3 } x + \frac { 2 } { 3 }$ 
G)
$\frac { 2 } { 9 } x - \frac { 7 } { 9 }$  
H) $- \frac { 1 } { 3 } x + \frac { 7 } { 3 }$

Quizplus · Accepted Answer

The linear approximation of a function $f(x)$ at a point $a$ is given by $L(x) = f(a) + f'(a)(x - a)$. First, we find $f(2)$ and $f'(2)$ for $f(x) = \frac{1}{\sqrt{x^3 + 1}}$.1. Calculate $f(2)$:$$f(2) = \frac{1}{\sqrt{2^3 + 1}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$$2. Calculate $f'(x)$:$$f'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{x^3 + 1}}\right) = \frac{d}{dx}\left(x^3 + 1\right)^{-\frac{1}{2}}$$Using the chain rule:$$f'(x) = -\frac{1}{2}(x^3 + 1)^{-\frac{3}{2}} \cdot 3x^2 = -\frac{3x^2}{2\sqrt{(x^3 + 1)^3}}$$3. Calculate $f'(2)$:$$f'(2) = -\frac{3 \cdot 2^2}{2\sqrt{(2^3 + 1)^3}} = -\frac{12}{2\sqrt{9^3}} = -\frac{12}{2 \cdot 27} = -\frac{12}{54} = -\frac{2}{9}$$4. Form the linear approximation $L(x)$:$$L(x) = f(2) + f'(2)(x - 2) = \frac{1}{3} - \frac{2}{9}(x - 2)$$$$L(x) = \frac{1}{3} - \frac{2}{9}x + \frac{4}{9} = -\frac{2}{9}x + \frac{7}{9}$$Therefore, the correct linear approximation is $-\frac{2}{9}x + \frac{7}{9}$, which corresponds to choice C.

Find the Linear Approximation To f(x)=1x3+1f ( x ) = \frac { 1 } { \sqrt { x ^ { 3 } + 1 } }f(x)=x3+1​1​

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