Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 2

Determine the Linearization $L(x)$ of the function $\displaystyle f(x) = \frac{1}{\sqrt{2 + x}}$ at $a = 0$.

Using the Linearization Formula

$L(x) = f(a) + f'(a)(x - a)$


$
\begin{equation}
\begin{aligned}

f(a) = f(0) =& \frac{1}{\sqrt{2 + 0}}
\\
\\
f(0) =& \frac{1}{\sqrt{2}}
\\
\\
f'(a) = f'(0) =& \frac{\displaystyle (2 + x)^{\frac{1}{2}} \frac{d}{dx} (1) - (1) \frac{d}{dx} (2 + x)^{\frac{1}{2}} }{[(2 + x)^{\frac{1}{2}}]^2}
\\
\\
f'(0) =& \frac{(2 + x)^{\frac{1}{2}} (0) - (1) \left( \frac{1}{2} \right) (2 + x)^{\frac{-1}{2}} \frac{d}{dx} (2 + x)}{2 + x}
\\
\\
f'(0) =& \frac{\displaystyle \frac{-1}{2} (2 + x)^{\frac{-1}{2}} (1)}{2 + x}
\\
\\
f'(0) =& \frac{-1}{2(2 + x)(2 + x)^{\frac{1}{2}}}
\\
\\
f'(0) =& \frac{-1}{2(2 + x)^{\frac{3}{2}}}
\\
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f'(0) =& \frac{-1}{2(2 + 0)^{\frac{3}{2}}}
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f'(0) =& \frac{-1}{2(2^3)^{\frac{1}{2}}} = \frac{-1}{2(8)^{\frac{1}{2}}} = \frac{-1}{2(4 \cdot 2)^{\frac{1}{2}}} = \frac{-1}{4(2)^{\frac{1}{2}}} = \frac{-1}{4 \sqrt{2}}
\\
\\
L(x) =& \frac{1}{\sqrt{2}} + \left( \frac{-1}{4 \sqrt{2}} \right) (x - 0)
\\
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L(x) =& \frac{1}{\sqrt{2}} - \frac{x}{4 \sqrt{2} }
\\
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L(x) =& \frac{1}{\sqrt{2}} \left( 1 - \frac{x}{4} \right)

\end{aligned}
\end{equation}
$