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Calculus
Applications of Derivatives
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Find the linear approximation of \(\displaystyle f(x) = \ln{\left(x + 1 \right)}\) at \(\displaystyle a = 0\) and use it to approximate \(\displaystyle \ln{0.83}\)

Using the formula for the linearization \(\displaystyle L(x) = f'(a)(x-a)+f(a)\) gives: \begin{equation*}f(x) = \log{\left(x + 1 \right)} \approx L(x) = x \end{equation*} To approximate \(\displaystyle \ln({0.83})\) use \(\displaystyle x = 0.83-1 = -0.17\). \begin{equation*}\ln({0.83}) = f(-0.17) \approx L(-0.17) = -0.17 \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the linear approximation of $f(x) = \ln{\left(x + 1 \right)}$ at $a = 0$ and use it to approximate $\ln{0.83}$
    \soln{9cm}{Using the formula for the linearization $L(x) = f'(a)(x-a)+f(a)$ gives:
\begin{equation*}f(x) = \log{\left(x + 1 \right)} \approx L(x) = x \end{equation*}
To approximate $\ln({0.83})$ use $x = 0.83-1 = -0.17$.
\begin{equation*}\ln({0.83}) = f(-0.17)  \approx L(-0.17) = -0.17 \end{equation*}
}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the linear approximation of  <img class="equation_image" title=" \displaystyle f(x) = \ln{\left(x + 1 \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cln%7B%5Cleft%28x%20%2B%201%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(x) = \ln{\left(x + 1 \right)} " data-equation-content=" \displaystyle f(x) = \ln{\left(x + 1 \right)} " />  at  <img class="equation_image" title=" \displaystyle a = 0 " src="/equation_images/%20%5Cdisplaystyle%20a%20%3D%200%20" alt="LaTeX:  \displaystyle a = 0 " data-equation-content=" \displaystyle a = 0 " />  and use it to approximate  <img class="equation_image" title=" \displaystyle \ln{0.83} " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%7B0.83%7D%20" alt="LaTeX:  \displaystyle \ln{0.83} " data-equation-content=" \displaystyle \ln{0.83} " /> </p> </p>
HTML for Canvas
<p> <p>Using the formula for the linearization  <img class="equation_image" title=" \displaystyle L(x) = f'(a)(x-a)+f(a) " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%20%3D%20f%27%28a%29%28x-a%29%2Bf%28a%29%20" alt="LaTeX:  \displaystyle L(x) = f'(a)(x-a)+f(a) " data-equation-content=" \displaystyle L(x) = f'(a)(x-a)+f(a) " />  gives:
 <img class="equation_image" title=" f(x) = \log{\left(x + 1 \right)} \approx L(x) = x  " src="/equation_images/%20f%28x%29%20%3D%20%5Clog%7B%5Cleft%28x%20%2B%201%20%5Cright%29%7D%20%5Capprox%20L%28x%29%20%3D%20x%20%20" alt="LaTeX:  f(x) = \log{\left(x + 1 \right)} \approx L(x) = x  " data-equation-content=" f(x) = \log{\left(x + 1 \right)} \approx L(x) = x  " /> 
To approximate  <img class="equation_image" title=" \displaystyle \ln({0.83}) " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%28%7B0.83%7D%29%20" alt="LaTeX:  \displaystyle \ln({0.83}) " data-equation-content=" \displaystyle \ln({0.83}) " />  use  <img class="equation_image" title=" \displaystyle x = 0.83-1 = -0.17 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%200.83-1%20%3D%20-0.17%20" alt="LaTeX:  \displaystyle x = 0.83-1 = -0.17 " data-equation-content=" \displaystyle x = 0.83-1 = -0.17 " /> .
 <img class="equation_image" title=" \ln({0.83}) = f(-0.17)  \approx L(-0.17) = -0.17  " src="/equation_images/%20%5Cln%28%7B0.83%7D%29%20%3D%20f%28-0.17%29%20%20%5Capprox%20L%28-0.17%29%20%3D%20-0.17%20%20" alt="LaTeX:  \ln({0.83}) = f(-0.17)  \approx L(-0.17) = -0.17  " data-equation-content=" \ln({0.83}) = f(-0.17)  \approx L(-0.17) = -0.17  " /> 
</p> </p>