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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)=exf ( x ) = e ^ { - x }

, a=ln 2

Solution

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

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

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

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

We have\textbf{We have}

f(ln2)=eln2=eln21=21=12f(ln2)=eln2=eln21=21=12\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*}

Our approximation\textbf{Our approximation} becomes

L(x)=1212(xln2)=1212x+ln22=12x+12+ln22\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*}

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

L(x)=12x+12+ln22\boxed{\color{#c34632}{L(x)=-\frac{1}{2}x+\frac{1}{2}+\frac{\ln{2}}{2}}}

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