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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 \(\displaystyle \ln{1.49}\)

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 \(\displaystyle \ln({1.49})\) use \(\displaystyle x = 1.49-1 = 0.49\). \begin{equation*}\ln({1.49}) = f(0.49) \approx L(0.49) = 0.49 \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 $\ln{1.49}$
    \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 $\ln({1.49})$ use $x = 1.49-1 = 0.49$.
\begin{equation*}\ln({1.49}) = f(0.49)  \approx L(0.49) = 0.49 \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{1.49} " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%7B1.49%7D%20" alt="LaTeX:  \displaystyle \ln{1.49} " data-equation-content=" \displaystyle \ln{1.49} " /> </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({1.49}) " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%28%7B1.49%7D%29%20" alt="LaTeX:  \displaystyle \ln({1.49}) " data-equation-content=" \displaystyle \ln({1.49}) " />  use  <img class="equation_image" title=" \displaystyle x = 1.49-1 = 0.49 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201.49-1%20%3D%200.49%20" alt="LaTeX:  \displaystyle x = 1.49-1 = 0.49 " data-equation-content=" \displaystyle x = 1.49-1 = 0.49 " /> .
 <img class="equation_image" title=" \ln({1.49}) = f(0.49)  \approx L(0.49) = 0.49  " src="/equation_images/%20%5Cln%28%7B1.49%7D%29%20%3D%20f%280.49%29%20%20%5Capprox%20L%280.49%29%20%3D%200.49%20%20" alt="LaTeX:  \ln({1.49}) = f(0.49)  \approx L(0.49) = 0.49  " data-equation-content=" \ln({1.49}) = f(0.49)  \approx L(0.49) = 0.49  " /> 
</p> </p>