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Find the linearization of \(\displaystyle f(x) = \sqrt{x}\) at \(\displaystyle x = 64\) and use it to approximate \(\displaystyle \sqrt{39}\). Leave your solution as a rational number.

The formula for the linearization of \(\displaystyle f\) at \(\displaystyle x=a\) is: \(\displaystyle L(x)=f'(a)(x-a)+f(a)\). \(\displaystyle f(64) = 8\) and caclulating the derivative gives \(\displaystyle f'(x) = \frac{1}{2 \sqrt{x}}\) and \(\displaystyle f'(64)=\frac{1}{16}\). The linearization is \(\displaystyle L(x)=\frac{1}{16} \left(x - 64\right) + 8 = \frac{x}{16} + 4\)This gives the approximation \(\displaystyle \sqrt{39} \approx L(39) = \frac{103}{16}\).

Download \(\LaTeX\) 

\begin{question}Find the linearization of $f(x) = \sqrt{x}$ at $x = 64$ and use it to approximate $\sqrt{39}$. Leave your solution as a rational number. 
    \soln{9cm}{The formula for the linearization of $f$ at $x=a$ is: $L(x)=f'(a)(x-a)+f(a)$. $f(64) = 8$ and caclulating the derivative gives $f'(x) = \frac{1}{2 \sqrt{x}}$ and $f'(64)=\frac{1}{16}$. The linearization is $L(x)=\frac{1}{16} \left(x - 64\right) + 8 = \frac{x}{16} + 4$This gives the approximation $\sqrt{39} \approx L(39) = \frac{103}{16}$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the linearization of  <img class="equation_image" title=" \displaystyle f(x) = \sqrt{x} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csqrt%7Bx%7D%20" alt="LaTeX:  \displaystyle f(x) = \sqrt{x} " data-equation-content=" \displaystyle f(x) = \sqrt{x} " />  at  <img class="equation_image" title=" \displaystyle x = 64 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%2064%20" alt="LaTeX:  \displaystyle x = 64 " data-equation-content=" \displaystyle x = 64 " />  and use it to approximate  <img class="equation_image" title=" \displaystyle \sqrt{39} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B39%7D%20" alt="LaTeX:  \displaystyle \sqrt{39} " data-equation-content=" \displaystyle \sqrt{39} " /> . Leave your solution as a rational number. </p> </p>
HTML for Canvas
<p> <p>The formula for the linearization of  <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX:  \displaystyle f " data-equation-content=" \displaystyle f " />  at  <img class="equation_image" title=" \displaystyle x=a " src="/equation_images/%20%5Cdisplaystyle%20x%3Da%20" alt="LaTeX:  \displaystyle x=a " data-equation-content=" \displaystyle x=a " />  is:  <img class="equation_image" title=" \displaystyle L(x)=f'(a)(x-a)+f(a) " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%3Df%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) " /> .  <img class="equation_image" title=" \displaystyle f(64) = 8 " src="/equation_images/%20%5Cdisplaystyle%20f%2864%29%20%3D%208%20" alt="LaTeX:  \displaystyle f(64) = 8 " data-equation-content=" \displaystyle f(64) = 8 " />  and caclulating the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B1%7D%7B2%20%5Csqrt%7Bx%7D%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " data-equation-content=" \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " />  and  <img class="equation_image" title=" \displaystyle f'(64)=\frac{1}{16} " src="/equation_images/%20%5Cdisplaystyle%20f%27%2864%29%3D%5Cfrac%7B1%7D%7B16%7D%20" alt="LaTeX:  \displaystyle f'(64)=\frac{1}{16} " data-equation-content=" \displaystyle f'(64)=\frac{1}{16} " /> . The linearization is  <img class="equation_image" title=" \displaystyle L(x)=\frac{1}{16} \left(x - 64\right) + 8 = \frac{x}{16} + 4 " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%3D%5Cfrac%7B1%7D%7B16%7D%20%5Cleft%28x%20-%2064%5Cright%29%20%2B%208%20%3D%20%5Cfrac%7Bx%7D%7B16%7D%20%2B%204%20" alt="LaTeX:  \displaystyle L(x)=\frac{1}{16} \left(x - 64\right) + 8 = \frac{x}{16} + 4 " data-equation-content=" \displaystyle L(x)=\frac{1}{16} \left(x - 64\right) + 8 = \frac{x}{16} + 4 " /> This gives the approximation  <img class="equation_image" title=" \displaystyle \sqrt{39} \approx L(39) = \frac{103}{16} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B39%7D%20%5Capprox%20L%2839%29%20%3D%20%5Cfrac%7B103%7D%7B16%7D%20" alt="LaTeX:  \displaystyle \sqrt{39} \approx L(39) = \frac{103}{16} " data-equation-content=" \displaystyle \sqrt{39} \approx L(39) = \frac{103}{16} " /> . </p> </p>