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Question

# Carry out the following steps for the given functions f and points a. Find the linear approximation L to the function f at the point a.$f ( x ) = e ^ { - x }$, a=ln 2

Solution

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Step 1
1 of 2

$\textbf{To find}$ linear approximation we will use the formula

$\color{#c34632}{L(x)=f(a)+f'(a)(x-a)}$

$\textbf{First thing}$ we will do is to find first derivative

\begin{align*} f(x)&=e^{-x}\quad \Big/\frac{d}{dx}\\ f'(x)&=-e^{-x}\\ \end{align*}

$\textbf{We have}$

\begin{align*} f(\ln{2})&=e^{-\ln{2}}\tag{a\ln{b}=\ln{b^a}}\\ &=e^{\ln{2^{-1}}}\\ &=2^{-1}\\ &=\frac{1}{2}\\ \\ f'(\ln{2})&=-e^{-\ln{2}}\\ &=-e^{\ln{2^{-1}}}\\ &=-2^{-1}\\ &=-\frac{1}{2} \end{align*}

$\textbf{Our approximation}$ becomes

\begin{align*} L(x)&=\frac{1}{2}-\frac{1}{2}(x-\ln{2})\\ &=\frac{1}{2}-\frac{1}{2}x+\frac{\ln{2}}{2}\\ &=-\frac{1}{2}x+\frac{1}{2}+\frac{\ln{2}}{2}\\ \end{align*}

$\textbf{Thus}$ our $\textbf{final solution}$ becomes

$\boxed{\color{#c34632}{L(x)=-\frac{1}{2}x+\frac{1}{2}+\frac{\ln{2}}{2}}}$

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