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Calculus
Applications of Derivatives
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Find the linear approximation of \(\displaystyle f(x) = \sqrt{x + 289}\) at \(\displaystyle a = 0\) and use it to approximate \(\displaystyle \sqrt{289.80}\)

Using the formula for the linearization \(\displaystyle L(x) = f'(a)(x-a)+f(a)\) gives: \begin{equation*}f(x) = \sqrt{x + 289} \approx L(x) = \frac{x}{34} + 17 \end{equation*} To approximate \(\displaystyle \sqrt{289.80}\) use \(\displaystyle x = 289.80-289 = 0.80\). \begin{equation*}\sqrt{289.80} = f(0.80) \approx L(0.80) = 17.023529 \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the linear approximation of $f(x) = \sqrt{x + 289}$ at $a = 0$ and use it to approximate $\sqrt{289.80}$
    \soln{9cm}{Using the formula for the linearization $L(x) = f'(a)(x-a)+f(a)$ gives:
\begin{equation*}f(x) = \sqrt{x + 289} \approx L(x) = \frac{x}{34} + 17 \end{equation*}
To approximate $\sqrt{289.80}$ use $x = 289.80-289 = 0.80$.
\begin{equation*}\sqrt{289.80} = f(0.80)  \approx L(0.80) = 17.023529 \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) = \sqrt{x + 289} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20289%7D%20" alt="LaTeX:  \displaystyle f(x) = \sqrt{x + 289} " data-equation-content=" \displaystyle f(x) = \sqrt{x + 289} " />  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 \sqrt{289.80} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B289.80%7D%20" alt="LaTeX:  \displaystyle \sqrt{289.80} " data-equation-content=" \displaystyle \sqrt{289.80} " /> </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) = \sqrt{x + 289} \approx L(x) = \frac{x}{34} + 17  " src="/equation_images/%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20289%7D%20%5Capprox%20L%28x%29%20%3D%20%5Cfrac%7Bx%7D%7B34%7D%20%2B%2017%20%20" alt="LaTeX:  f(x) = \sqrt{x + 289} \approx L(x) = \frac{x}{34} + 17  " data-equation-content=" f(x) = \sqrt{x + 289} \approx L(x) = \frac{x}{34} + 17  " /> 
To approximate  <img class="equation_image" title=" \displaystyle \sqrt{289.80} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B289.80%7D%20" alt="LaTeX:  \displaystyle \sqrt{289.80} " data-equation-content=" \displaystyle \sqrt{289.80} " />  use  <img class="equation_image" title=" \displaystyle x = 289.80-289 = 0.80 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20289.80-289%20%3D%200.80%20" alt="LaTeX:  \displaystyle x = 289.80-289 = 0.80 " data-equation-content=" \displaystyle x = 289.80-289 = 0.80 " /> .
 <img class="equation_image" title=" \sqrt{289.80} = f(0.80)  \approx L(0.80) = 17.023529  " src="/equation_images/%20%5Csqrt%7B289.80%7D%20%3D%20f%280.80%29%20%20%5Capprox%20L%280.80%29%20%3D%2017.023529%20%20" alt="LaTeX:  \sqrt{289.80} = f(0.80)  \approx L(0.80) = 17.023529  " data-equation-content=" \sqrt{289.80} = f(0.80)  \approx L(0.80) = 17.023529  " /> 
</p> </p>