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Calculus
Applications of Derivatives
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Find the linear approximation of \(\displaystyle f(x) = \sqrt{x + 169}\) at \(\displaystyle a = 0\) and use it to approximate \(\displaystyle \sqrt{169.08}\)

Using the formula for the linearization \(\displaystyle L(x) = f'(a)(x-a)+f(a)\) gives: \begin{equation*}f(x) = \sqrt{x + 169} \approx L(x) = \frac{x}{26} + 13 \end{equation*} To approximate \(\displaystyle \sqrt{169.08}\) use \(\displaystyle x = 169.08-169 = 0.08\). \begin{equation*}\sqrt{169.08} = f(0.08) \approx L(0.08) = 13.003077 \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the linear approximation of $f(x) = \sqrt{x + 169}$ at $a = 0$ and use it to approximate $\sqrt{169.08}$
    \soln{9cm}{Using the formula for the linearization $L(x) = f'(a)(x-a)+f(a)$ gives:
\begin{equation*}f(x) = \sqrt{x + 169} \approx L(x) = \frac{x}{26} + 13 \end{equation*}
To approximate $\sqrt{169.08}$ use $x = 169.08-169 = 0.08$.
\begin{equation*}\sqrt{169.08} = f(0.08)  \approx L(0.08) = 13.003077 \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 + 169} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20169%7D%20" alt="LaTeX:  \displaystyle f(x) = \sqrt{x + 169} " data-equation-content=" \displaystyle f(x) = \sqrt{x + 169} " />  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{169.08} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B169.08%7D%20" alt="LaTeX:  \displaystyle \sqrt{169.08} " data-equation-content=" \displaystyle \sqrt{169.08} " /> </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 + 169} \approx L(x) = \frac{x}{26} + 13  " src="/equation_images/%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20169%7D%20%5Capprox%20L%28x%29%20%3D%20%5Cfrac%7Bx%7D%7B26%7D%20%2B%2013%20%20" alt="LaTeX:  f(x) = \sqrt{x + 169} \approx L(x) = \frac{x}{26} + 13  " data-equation-content=" f(x) = \sqrt{x + 169} \approx L(x) = \frac{x}{26} + 13  " /> 
To approximate  <img class="equation_image" title=" \displaystyle \sqrt{169.08} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B169.08%7D%20" alt="LaTeX:  \displaystyle \sqrt{169.08} " data-equation-content=" \displaystyle \sqrt{169.08} " />  use  <img class="equation_image" title=" \displaystyle x = 169.08-169 = 0.08 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20169.08-169%20%3D%200.08%20" alt="LaTeX:  \displaystyle x = 169.08-169 = 0.08 " data-equation-content=" \displaystyle x = 169.08-169 = 0.08 " /> .
 <img class="equation_image" title=" \sqrt{169.08} = f(0.08)  \approx L(0.08) = 13.003077  " src="/equation_images/%20%5Csqrt%7B169.08%7D%20%3D%20f%280.08%29%20%20%5Capprox%20L%280.08%29%20%3D%2013.003077%20%20" alt="LaTeX:  \sqrt{169.08} = f(0.08)  \approx L(0.08) = 13.003077  " data-equation-content=" \sqrt{169.08} = f(0.08)  \approx L(0.08) = 13.003077  " /> 
</p> </p>